Exercise 11.1: This exercise introduces the concept of conic sections, including the definitions and standard equations of the circle, parabola, ellipse, and hyperbola. Students will learn how to identify the type of conic section given its equation and solve problems related to it.

Exercise 11.2: In this exercise, students will learn about the various properties of the circle, such as its equation in different forms, its radius, center, and diameter, and the equation of a tangent to the circle.

Exercise 11.3: This exercise focuses on the properties of the parabola, including its standard equation, focus, vertex, axis, directrix, and latus rectum. Students will learn how to find the equation of a tangent to the parabola and solve problems related to it.

Exercise 11.4: This exercise covers the properties of the ellipse and hyperbola, including their standard equations, foci, vertices, axes, directrices, and eccentricity. Students will learn how to find the equation of a tangent to the ellipse or hyperbola and solve problems related to them.

Miscellaneous Exercise: This exercise includes a mix of questions covering all the concepts taught in the chapter. Students will have to apply their knowledge of conic sections to solve various problems and answer questions. They will also practice finding the equation of a tangent to the circle, parabola, ellipse, and hyperbola.

Access NCERT Solutions for Class 11 Maths  Chapter 11 – Conic Sections

Exercise 11.1

1. Find the equation of the circle with centre \[(0,2)\] and radius \[2\]

Ans: The equation of a circle with centre \[(h,k)\] and radius \[r\] is given as

\[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\]

It is given that centre \[(h,k)=(0,2)\] and radius \[(r)=2.\]

Therefore, the equation of the circle is

\[{{(x-0)}^{2}}+{{(y-2)}^{2}}={{2}^{2}}\]

\[\begin{align}   & {{x}^{2}}+{{y}^{2}}+4-4y=4 \\  & {{x}^{2}}+{{y}^{2}}-4y=0 \\ \end{align}\]

2. Find the equation of the circle with centre \[(-2,3)\] and radius \[4\]

Ans: The equation of a circle with centre \[(h,k)\] and radius \[r\] is given as

\[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\]

It is given that centre \[(h,k)=(-2,3)\] and radius \[(r)=4.\]

Therefore, the equation of the circle is

\[{{(x+2)}^{2}}+{{(y-3)}^{2}}={{(4)}^{2}}\]

\[\begin{align}  & {{x}^{2}}+4x+4+{{y}^{2}}-6y+9=16 \\  & {{x}^{2}}+{{y}^{2}}+4x-6y-3=0 \\ \end{align}\]

3. Find the equation of the circle with centre \[\left( \dfrac{1}{2},\dfrac{1}{4} \right)\] and radius \[\left( \dfrac{1}{12} \right)\]

Ans: The equation of a circle with centre \[(h,k)\] and radius \[r\] is given as

\[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\]

It is given that centre \[(h,k)=\left( \dfrac{1}{2},\dfrac{1}{4} \right)\] and radius \[(r)=\left( \dfrac{1}{12} \right).\]

Therefore, the equation of the circle is

\[{{\left( x-\dfrac{1}{2} \right)}^{2}}+{{\left( y-\dfrac{1}{4} \right)}^{2}}={{\left( \dfrac{1}{12} \right)}^{2}}\]

\[{{x}^{2}}-x+\dfrac{1}{4}+{{y}^{2}}-\dfrac{y}{2}+\dfrac{1}{16}=\dfrac{1}{144}\]

\[{{x}^{2}}-x+\dfrac{1}{4}+{{y}^{2}}-\dfrac{y}{2}+\dfrac{1}{16}-\dfrac{1}{144}=0\]

\[144{{x}^{2}}-144x+36+144{{y}^{2}}-72y+9-1=0\] (Solve by taking LCM)

\[144{{x}^{2}}-144x+144{{y}^{2}}-72y+44=0\]

\[36{{x}^{2}}-36x+36{{y}^{2}}-18y+11=0\]

\[36{{x}^{2}}+36{{y}^{2}}-36x-18y+11=0\]

4. Find the equation of the circle with centre \[(1,1)\] and radius \[\sqrt{2}\]

Ans: The equation of a circle with centre \[(h,k)\] and radius \[r\] is given as

\[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\]

It is given that centre \[(h,k)=(1,1)\] and radius \[(r)=\sqrt{2}.\]

Therefore, the equation of the circle is

\[{{(x-1)}^{2}}+{{(y-1)}^{2}}={{(\sqrt{2})}^{2}}\]

\[\begin{align}  & {{x}^{2}}-2x+1+{{y}^{2}}-2y+1=2 \\  & {{x}^{2}}+{{y}^{2}}-2x-2y=0 \\ \end{align}\]

5. Find the equation of the circle with centre \[(-a,-b)\] and radius \[\sqrt{{{a}^{2}}-{{b}^{2}}}\]

Ans: The equation of a circle with centre \[(h,k)\] and radius \[r\] is given as

\[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\]

It is given that centre \[(h,k)=(-a,-b)\] and radius \[(r)=\sqrt{{{a}^{2}}-{{b}^{2}}}.\]

Therefore, the equation of the circle is

\[{{(x+a)}^{2}}+{{(y+b)}^{2}}={{(\sqrt{{{a}^{2}}-{{b}^{2}}})}^{2}}\]

\[\begin{align}   & {{x}^{2}}+2ax+{{a}^{2}}+{{y}^{2}}+2by+{{b}^{2}}={{a}^{2}}-{{b}^{2}} \\  & {{x}^{2}}+{{y}^{2}}+2ax+2by+2{{b}^{2}}=0 \\ \end{align}\]

6. Find the centre and radius of the circle \[{{(x+5)}^{2}}+{{(y-3)}^{2}}=36\]

Ans: The equation of the given circle is \[{{(x+5)}^{2}}+{{(y-3)}^{2}}=36\]

\[{{(x+5)}^{2}}+{{(y-3)}^{2}}=36\]

\[\Rightarrow {{\{x-(-5)\}}^{2}}+{{(y-3)}^{2}}={{(6)}^{2}}\], which is the form of \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\], where \[h=-5,k=3,and\; r=6\].

Thus, the centre of the given circle is \[(-5,3),\] while its radius is 6.

7. Find the centre and radius of the circle \[{{x}^{2}}+{{y}^{2}}-4x-8y-45=0\]

Ans: The equation of the given circle is \[{{x}^{2}}+{{y}^{2}}-4x-8y-45=0\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}-4x-8y-45=0\]

\[\Rightarrow ({{x}^{2}}-4x)+({{y}^{2}}-8y)=45\]

\[\Rightarrow \{{{x}^{2}}-2(x)(2)+{{(2)}^{2}}\}+\{{{y}^{2}}-2(y)(4)+{{(4)}^{2}}\}-4-16=45\]

\[\Rightarrow {{(x-2)}^{2}}+{{(y-4)}^{2}}=65\]

\[\Rightarrow {{(x-2)}^{2}}+{{(y-4)}^{2}}={{\left( \sqrt{65} \right)}^{2}},\] which is of the form \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\], where \[h=2,k=4,and\; r=\sqrt{65}\]

Thus, the centre of the given circle is \[(2,4),\]while its radius is \[\sqrt{65}\].

8. Find the centre and radius of the circle \[{{x}^{2}}+{{y}^{2}}-8x+10y-12=0\]

Ans: The equation of the given circle is \[{{x}^{2}}+{{y}^{2}}-8x+10y-12=0\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}-8x+10y-12=0\]
\[\Rightarrow ({{x}^{2}}-8x)+({{y}^{2}}+10y)=12\]

\[\Rightarrow \{{{x}^{2}}-2(x)(4)+{{(4)}^{2}}\}+\{{{y}^{2}}+2(y)(5)+{{(5)}^{2}}-16-25=12\]

\[\Rightarrow {{(x-4)}^{2}}+{{(y+5)}^{2}}=53\]

\[\Rightarrow {{(x-4)}^{2}}+{{\{y-(-5)\}}^{2}}={{\left( \sqrt{53} \right)}^{2}},\] which is of the form \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\], where \[h=4,k=-5,and\; r=\sqrt{53}\]

Thus, the centre of the given circle is \[(4,-5),\] while its radius is \[\sqrt{53}\].

9. Find the centre and radius of the circle \[2{{x}^{2}}+2{{y}^{2}}-x=0\]

Ans: The equation of the given circle is \[2{{x}^{2}}+2{{y}^{2}}-x=0\]

\[\Rightarrow 2{{x}^{2}}+2{{y}^{2}}-x=0\]
\[\Rightarrow (2{{x}^{2}}-x)+2{{y}^{2}}=0\]

\[\Rightarrow 2\left[ \left( {{x}^{2}}-\dfrac{x}{2} \right)+{{y}^{2}} \right]=0\]

\[\Rightarrow \left\{ {{x}^{2}}-2.x\left( \dfrac{1}{4} \right)+{{\left( \dfrac{1}{4} \right)}^{2}} \right\}+{{y}^{2}}-{{\left( \dfrac{1}{4} \right)}^{2}}=0\]

\[\Rightarrow {{\left( x-\dfrac{1}{4} \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( \dfrac{1}{4} \right)}^{2}},\] which is of the form \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\], where \[h=\dfrac{1}{4},k=0,\] and \[r = \dfrac{1}{4}\]

Thus, the centre of the given circle is \[\left( \dfrac{1}{4},0 \right),\] while its radius is \[\dfrac{1}{4}\].

10. Find the equation of the circle passing through the points \[(4,1)\] and \[(6,5)\] and whose centre is on the line \[4x+y=16\]

Ans: Let the equation of the required circle be \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\]

Since the circle passes through points \[(4,1)\] and \[(6,5)\],

\[{{(4-h)}^{2}}+{{(1-k)}^{2}}={{r}^{2}}\] ………(i)

\[{{(6-h)}^{2}}+{{(5-k)}^{2}}={{r}^{2}}\] ………(ii)

Since the centre (h,k) of the circle lies on line \[4x+y=16,\]

\[4h+k=16\] ………(iii)

From equations (i) and (ii), we get

\[\Rightarrow {{(4-h)}^{2}}+{{(1-k)}^{2}}={{(6-h)}^{2}}+{{(5-k)}^{2}}\]

\[\Rightarrow 16-8h+{{h}^{2}}+1-2k+{{k}^{2}}=36-12h+{{h}^{2}}+25-10k+{{k}^{2}}\]

\[\begin{align}   & \Rightarrow 16-8h+1-2k=36-12h+25-10k \\  & \Rightarrow 4h+8k=44 \\ \end{align}\]

\[\Rightarrow h+2k=11\] ………(iv)

On solving equations (iii) and (iv), we obtain \[h=3\] and \[k=4\]

On substituting the values of h and k in equation (i), we obtain

\[{{(4-3)}^{2}}+{{(1-4)}^{2}}={{r}^{2}}\]

\[\Rightarrow {{(1)}^{2}}+{{(-3)}^{2}}={{r}^{2}}\]

\[\Rightarrow 1+9={{r}^{2}}\]

\[\begin{align} & \Rightarrow {{r}^{2}}=10 \\  & \Rightarrow r=\sqrt{10} \\ \end{align}\]

Thus, the equation of the required circle is

\[{{(x-3)}^{2}}+{{(y-4)}^{2}}={{\left( \sqrt{10} \right)}^{2}}\]

\[\begin{align}   & {{x}^{2}}-6x+9+{{y}^{2}}-8y+16=10 \\  & {{x}^{2}}+{{y}^{2}}-6x-8y+15=0 \\ \end{align}\]

11. Find the equation of the circle passing through the points \[(2,3)\] and \[(-1,1)\] and whose centre is on the line \[x-3y-11=0\]

Ans: Let the equation of the required circle be \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\]

Since the circle passes through points \[(2,3)\] and \[(-1,1)\],

\[{{(2-h)}^{2}}+{{(3-k)}^{2}}={{r}^{2}}\] ………(i)

\[{{(-1-h)}^{2}}+{{(1-k)}^{2}}={{r}^{2}}\] ………(ii)

Since the centre (h,k) of the circle lies on line \[x-3y-11=0\]

\[h-3k=11\] ………(iii)

From equations (i) and (ii), we get

\[\Rightarrow {{(2-h)}^{2}}+{{(3-k)}^{2}}={{(-1-h)}^{2}}+{{(1-k)}^{2}}\]

\[\Rightarrow 4-4h+{{h}^{2}}+9-6k+{{k}^{2}}=1+2h+{{h}^{2}}+1-2k+{{k}^{2}}\]

\[\Rightarrow 4-4h+9-6k=1+2h+1-2k\]

\[\Rightarrow 6h+4k=11\] ………(iv)

On solving equations (iii) and (iv), we obtain \[h=\dfrac{7}{2}\] and \[k=\dfrac{-5}{2}\]

On substituting the values of h and k in equation (i), we obtain

\[\Rightarrow {{\left( 2-\dfrac{7}{2} \right)}^{2}}+{{\left( 3+\dfrac{5}{2} \right)}^{2}}={{r}^{2}}\]

\[\Rightarrow {{\left( \dfrac{4-7}{2} \right)}^{2}}+{{\left( \dfrac{6+5}{2} \right)}^{2}}={{r}^{2}}\]

\[\Rightarrow {{\left( \dfrac{-3}{2} \right)}^{2}}+{{\left( \dfrac{11}{2} \right)}^{2}}={{r}^{2}}\]

\[\begin{align}   & \Rightarrow \dfrac{9}{4}+\dfrac{121}{4}={{r}^{2}} \\  & \Rightarrow \dfrac{130}{4}={{r}^{2}} \\ \end{align}\]

Thus, the equation of the required circle is

\[{{\left( x-\dfrac{7}{2} \right)}^{2}}+{{\left( y+\dfrac{5}{2} \right)}^{2}}=\dfrac{130}{4}\]

\[\begin{align}   & {{\left( \dfrac{2x-7}{2} \right)}^{2}}+{{\left( \dfrac{2y+5}{2} \right)}^{2}}=\dfrac{130}{4} \\ & 4{{x}^{2}}-28x+49+4{{y}^{2}}+20y+25=130 \\ \end{align}\]

\[\begin{align}   & 4{{x}^{2}}+4{{y}^{2}}-28x+20y-56=0 \\  & 4({{x}^{2}}+y{}^{2}-7x+5y-14)=0 \\ \end{align}\]

\[{{x}^{2}}+{{y}^{2}}-7x+5y-14=0\]

12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point \[(2,3)\].

Ans: Let the equation of the required circle be \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\]

Since the radius of the circle  is 5 and its centre lies on the x-axis, \[k=0\] and \[r=5\].

Now, the equation of the circle passes through point \[(2,3).\]

\[\begin{align}   & \therefore {{\left( 2-h \right)}^{2}}+{{3}^{2}}=25 \\  & \Rightarrow {{\left( 2-h \right)}^{2}}=25-9 \\  & \Rightarrow {{\left( 2-h \right)}^{2}}=16 \\ \end{align}\]

\[\begin{align}  & \Rightarrow \left( 2-h \right)=\pm \sqrt{16} \\  & =\pm 4 \\ \end{align}\]

If \[2-h=4,\] then \[h=-2\]

If \[2-h=-4,\] then \[h=6\]

When \[h=-2\], the equation of the circle becomes

\[\begin{align}  & {{\left( x+2 \right)}^{2}}+{{y}^{2}}=25 \\  & {{x}^{2}}+4x+4+{{y}^{2}}=25 \\  & {{x}^{2}}+{{y}^{2}}+4x-21=0 \\ \end{align}\]

When \[h=6\], the equation of the circle becomes

\[\begin{align}   & {{\left( x-6 \right)}^{2}}+{{y}^{2}}=25 \\  & {{x}^{2}}-12x+36+{{y}^{2}}=25 \\  & {{x}^{2}}+{{y}^{2}}-12x+11=0 \\ \end{align}\] So, the equation of the circle can be \[{{x}^{2}}+{{y}^{2}}+4x-21=0 \] or \[{{x}^{2}}+{{y}^{2}}-12x+11=0 \]

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13. Find the equation of the circle passing through \[(0,0)\] and making intercepts a and b on the coordinate axes.

Ans: Let the equation of the required circle be \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\]

Since the circle passes through \[(0,0)\].

\[\begin{align}  & {{(0-h)}^{2}}+{{(0-k)}^{2}}={{r}^{2}} \\  & \Rightarrow {{h}^{2}}+{{k}^{2}}={{r}^{2}} \\ \end{align}\]

The equation of the circle now becomes \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{h}^{2}}+{{k}^{2}}\]

It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points \[(a,0)\] and \[(0,b)\]. Therefore,

\[{{(a-h)}^{2}}+{{(0-k)}^{2}}={{h}^{2}}+{{k}^{2}}\] …….(1)

\[{{(0-h)}^{2}}+{{(b-k)}^{2}}={{h}^{2}}+{{k}^{2}}\] …….(2)

From equation (1), we’ll get

\[\begin{align}  & {{a}^{2}}-2ah+{{h}^{2}}+{{k}^{2}}={{h}^{2}}+{{k}^{2}} \\  & \Rightarrow {{a}^{2}}-2ah=0 \\  & \Rightarrow a(a-2h)=0 \\ \end{align}\]

\[\Rightarrow a=0\] or \[(a-2h)=0\]

However, \[a\ne 0;\] hence, \[(a-2h)=0\Rightarrow h=\dfrac{a}{2}\]

From equation (2), we’ll get

\[\begin{align}  & {{h}^{2}}+{{b}^{2}}-2bk+{{k}^{2}}={{h}^{2}}+{{k}^{2}} \\  & \Rightarrow {{b}^{2}}-2bk=0 \\  & \Rightarrow b(b-2k)=0 \\ \end{align}\]

\[\Rightarrow b=0\] or \[(b-2k)=0\]

However, \[b\ne 0;\] hence, \[(b-2k)=0\Rightarrow k=\dfrac{b}{2}\]

Thus, the equation of the required circle is

\[{{\left( x-\dfrac{a}{2} \right)}^{2}}+{{\left( y-\dfrac{b}{2} \right)}^{2}}={{\left( \dfrac{a}{2} \right)}^{2}}+{{\left( \dfrac{b}{2} \right)}^{2}}\]

\[{{\left( \dfrac{2x-a}{2} \right)}^{2}}+{{\left( \dfrac{2y-b}{2} \right)}^{2}}=\dfrac{{{a}^{2}}+{{b}^{2}}}{4}\]

\[\begin{align}  & \Rightarrow 4{{x}^{2}}-4ax+{{a}^{2}}+4{{y}^{2}}-4by+{{b}^{2}}={{a}^{2}}+{{b}^{2}} \\  & \Rightarrow 4{{x}^{2}}+4{{y}^{2}}-4ax-4by=0 \\  & \Rightarrow {{x}^{2}}+{{y}^{2}}-ax-by=0 \\ \end{align}\]

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14. Find the equation of a circle with centre \[(2,2)\] and passes through the point \[(4,5)\].

Ans: The centre of the circle is given as \[(h,k)=(2,2)\]

Since the circle passes through point \[(4,5),\] the radius \[(r)\] of the circle is the distance between the points \[(2,2)\] and \[(4,5)\].

\[\begin{align}  & \therefore r=\sqrt{{{(2-4)}^{2}}+{{(2-5)}^{2}}} \\  & =\sqrt{{{(-2)}^{2}}+{{(-3)}^{2}}} \\  & =\sqrt{4+9} \\  & =\sqrt{13} \\ \end{align}\]

Thus, the equation of the circle is 

\[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\]

\[\begin{align}  & {{(x-2)}^{2}}+{{(y-2)}^{2}}={{\left( \sqrt{13} \right)}^{2}} \\  & {{x}^{2}}-4x+4+{{y}^{2}}-4y+4=13 \\  & {{x}^{2}}+{{y}^{2}}-4x-4y-5=0 \\ \end{align}\]

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15. Does the point \[(-2.5,3.5)\] lie inside, outside or on the circle \[{{x}^{2}}+{{y}^{2}}=25\]?

Ans: The equation of the given circle is \[{{x}^{2}}+{{y}^{2}}=25\].

\[{{x}^{2}}+{{y}^{2}}=25\]

\[\Rightarrow {{(x-0)}^{2}}+{{(y-0)}^{2}}={{5}^{2}},\] which is of the form of \[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\], where \[h = 0, k = 0 \; and\; r=5\]

\[\therefore \] Centre \[=(0,0)\] and radius \[=5\]

Distance between point \[(-2.5,3.5)\] and center \[(0,0)\]

\[=\sqrt{{{(-2.5-0)}^{2}}+{{(3.5-0)}^{2}}}\]

\[\begin{align}  & =\sqrt{6.25+12.25} \\  & =\sqrt{18.25} \\ \end{align}\]

\[=4.3\] (approx.) \[<5\]

Since the distance between point \[(-2.5,3.5)\] and center \[(0,0)\] of the circle is less than the radius of the circle, point \[(-2.5,3.5)\] lies inside the circle.

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Exercise 11.2

1. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for \[{{y}^{2}}=12x\]

Ans: The given equation is \[{{y}^{2}}=12x\].

Here, the coefficient of x is positive. Hence, the parabola opens towards the right. On comparing this equation with \[{{y}^{2}}=4ax,\] we’ll get

\[4a=12\Rightarrow a=3\]

\[\therefore \] Coordinates of the focus = \[=(a,0)=(3,0)\]

Since the given equation involves \[{{y}^{2}}\], the axis of the parabola is the x-axis. Equation of directrix, \[x=-a\text{  }i.e.,\text{  }x=-3\text{  }i.e.,\text{  }x+3=0\] 

Length of latus rectum \[=4a=4\times 3=12\]

2. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for \[{{x}^{2}}=6y\]

Ans: The given equation is \[{{x}^{2}}=6y\]. 

Here, the coefficient of y is positive. Hence, the parabola opens upwards. 

On comparing this equation with \[{{x}^{2}}=4ay\] we obtain 

\[4a=6\Rightarrow a=\dfrac{3}{2}\]

\[\therefore \]Coordinates of the focus \[=(0,a)=\left( 0,\dfrac{3}{2} \right)\]

Since the given equation involves \[{{x}^{2}}\], the axis of the parabola is the y-axis. Equation of directrix, \[y=-a\]  i.e., \[y=\dfrac{-3}{2}\]

Length of latus rectum \[=4a=6\]

3. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for \[{{y}^{2}}=-8x\]

Ans: The given equation is \[{{y}^{2}}=-8x\].

Here, the coefficient of \[x\] is negative. Hence, the parabola opens towards the left. On comparing this equation with \[{{y}^{2}}=-4ax,\] we’ll get

\[-4a=-8\Rightarrow a=2\]

\[\therefore \] Coordinates of the focus \[=(-a,0)=(-2,0)\]

Since the given equation involves \[{{y}^{2}}\], the axis of the parabola is the x-axis. Equation of directrix, \[x=a\]  i.e., \[x=2\]

Length of latus rectum \[=4a=8\]

4. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for \[{{x}^{2}}=-16y\]

Ans: The given equation is \[{{x}^{2}}=-16y.\] 

Here, the coefficient of \[y\] is negative. Hence, the parabola opens downwards. 

On comparing this equation with \[{{x}^{2}}=-4ay\], we’ll get

\[-4a=-16\Rightarrow a=4\]

\[\therefore \] Coordinates of the focus \[=(0,-a)=(0,-4)\]

Since the given equation involves \[{{x}^{2}}\], the axis of the parabola is the y-axis. Equation of directrix, \[y=a\]  i.e., \[y=4\]

Length of latus rectum \[=4a=16\]

5. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for \[{{y}^{2}}=10x\]

Ans:The given equation is \[{{y}^{2}}=10x\].

Here, the coefficient of \[x\] is positive. Hence, the parabola opens towards the right. On comparing this equation with \[{{y}^{2}}=4ax\], we’ll get

\[4a=10\Rightarrow a=\dfrac{5}{2}\]

\[\therefore \] Coordinates of the focus \[=(a,0)=\left( \dfrac{5}{2},0 \right)\]

Since the given equation involves \[{{y}^{2}}\], the axis of the parabola is the x-axis. Equation of directrix, \[x=-a\]  i.e., \[x=-\dfrac{5}{2}\]

Length of latus rectum \[=4a=10\]

6. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for \[{{x}^{2}}=-9y\]

Ans: The given equation is \[{{x}^{2}}=-9y\]. 

Here, the coefficient of \[y\] is negative. Hence, the parabola opens downwards. 

On comparing this equation with \[{{x}^{2}}=-4ay\], we’ll get

\[-4a=-9\Rightarrow a=\dfrac{9}{4}\]

\[\therefore \] Coordinates of the focus \[=(0,-a)=\left( 0,-\dfrac{9}{4} \right)\]

Since the given equation involves \[{{x}^{2}}\], the axis of the parabola is the y-axis. Equation of directrix, \[y=a\]  i.e., \[y=\dfrac{9}{4}\]

Length of latus rectum \[=4a=9\]

7. Find the equation of the parabola that satisfies the following conditions: Focus \[(6,0);\] directrix \[x=-6\]

Ans: Focus \[(6,0);\]directrix, \[x=-6\]

Since the focus lies on the x-axis, the x-axis is the axis of the parabola. 

Therefore, the equation of the parabola is either of the form \[{{y}^{2}}=4ax\] or  \[{{y}^{2}}=-4ax\]. 

It is also seen that the directrix, \[x=-6\]is to the left of the y-axis, while the focus \[(6,0)\] is to the right of the y-axis. 

Hence, the parabola is of the form \[{{y}^{2}}=4ax\]. 

Here, \[a=6\]

Thus, the equation of the parabola is \[{{y}^{2}}=24x\].

8. Find the equation of the parabola that satisfies the following conditions: Focus \[(0,-3);\] directrix \[y=3\]

Ans: Focus \[=(0,-3);\] directrix \[y=3\]

Since the focus lies on the y-axis, the y-axis is the axis of the parabola. 

Therefore, the equation of the parabola is either of the form \[{{x}^{2}}=4ay\] or \[{{x}^{2}}=-4ay.\]

It is also seen that the directrix, \[y=3\] is above the x-axis, while the focus \[(0,-3)\] is below the x-axis. 

Hence, the parabola is of the form \[{{x}^{2}}=-4ay.\]

Here, \[a=3\]

Thus, the equation of the parabola is \[{{x}^{2}}=-12y.\] 

9. Find the equation of the parabola that satisfies the following conditions: Vertex \[(0,0);\] focus \[(3,0)\] 

Ans: Vertex \[(0,0);\] focus \[(3,0)\] 

Since the vertex of the parabola is \[(0,0)\] and the focus lies on the positive x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form \[{{y}^{2}}=4ax.\] 

Since the focus is \[(3,0)\], \[a=3\]. 

Thus, the equation of the parabola is \[{{y}^{2}}=4\times 3\times x\]  i.e., \[{{y}^{2}}=12x\] 

10. Find the equation of the parabola that satisfies the following conditions: Vertex \[(0,0)\] focus \[(-2,0)\]

Ans: Solution 10: Vertex \[(0,0)\] focus \[(-2,0)\] 

Since the vertex of the parabola is \[(0,0)\] and the focus lies on the negative x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form \[{{y}^{2}}=-4ax.\] 

Since the focus is \[(-2,0),\]\[a=2.\] 

Thus, the equation of the parabola is \[{{y}^{2}}=-4\times 2\times x\]  i.e., \[{{y}^{2}}=-8x\]

11. Find the equation of the parabola that satisfies the following conditions: Vertex \[(0,0)\]passing through \[(2,3)\]and axis is along x-axis 

Ans: Since the vertex is (0, 0) and the axis of the parabola is the x-axis, the equation of the parabola is either of the form \[{{y}^{2}}=4ax\] or \[{{y}^{2}}=-4ax.\]

The parabola passes through point \[(2,3)\], which lies in the first quadrant. Therefore, the equation of the parabola is of the form \[{{y}^{2}}=4ax\], while point \[(2,3)\] must satisfy the equation \[{{y}^{2}}=4ax\].

\[\therefore {{(3)}^{2}}=4a(2)\Rightarrow a=\dfrac{9}{8}\]

Thus, the equation of the parabola is \[{{y}^{2}}=4\left( \dfrac{9}{8} \right)x\]

\[\begin{align}   & \Rightarrow {{y}^{2}}=\dfrac{9}{2}x \\  & \Rightarrow 2{{y}^{2}}=9x \\ \end{align}\]

(image will be uploaded soon)

12. Find the equation of the parabola that satisfies the following conditions: Vertex \[(0,0)\], passing through \[(5,2)\] and symmetric with respect to y-axis

Ans: Since the vertex is \[(0,0)\] and the parabola is symmetric about the y-axis, the equation of the parabola is either of the form \[{{x}^{2}}=4ay\] or \[{{x}^{2}}=-4ay.\] 

The parabola passes through point \[(5,2)\], which lies in the first quadrant. Therefore, the equation of the parabola is of the form \[{{x}^{2}}=4ay\], while point \[(5,2)\] must satisfy the equation \[{{x}^{2}}=4ay\]. 

\[\begin{align}   & \therefore {{(5)}^{2}}=4\times a\times 2 \\  & \Rightarrow 25=8a \\  & \Rightarrow a=\dfrac{25}{8} \\ \end{align}\]

Thus, the equation of the parabola is

\[\Rightarrow {{x}^{2}}=4\left( \dfrac{25}{8} \right)y\]

\[\Rightarrow 2{{x}^{2}}=25y\]

(image will be uploaded soon)

Exercise 11.3

1. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \[\dfrac{{{x}^{2}}}{36}+\dfrac{{{y}^{2}}}{16}=1\] 

Ans: The given equation is \[\dfrac{{{x}^{2}}}{36}+\dfrac{{{y}^{2}}}{16}=1\]

Here, the denominator of \[\dfrac{{{x}^{2}}}{36}\] is greater than the denominator of \[\dfrac{{{y}^{2}}}{16}\]. 

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis. 

On comparing the given equation with \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\], we’ll get \[a=6\] and \[b=4\]

\[\begin{align}  & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\  & =\sqrt{36-16} \\  & =\sqrt{20}=2\sqrt{5} \\ \end{align}\]

Therefore, 

The coordinates of the foci are \[\left( 2\sqrt{5},0 \right)\] and \[\left( -2\sqrt{5},0 \right)\]

The coordinates of the vertices are \[\left( 6,0 \right)\]and\[\left( -6,0 \right)\].

Length of major axis \[=2a=12\]

Length of minor axis \[=2b=8\]

Eccentricity, \[e=\dfrac{c}{a}=\dfrac{2\sqrt{5}}{6}=\dfrac{\sqrt{5}}{3}\]

Length of latus rectum \[=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 16}{6}=\dfrac{16}{3}\]

2. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \[\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{25}=1\] 

Ans: The given equation is \[\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{25}=1\] or \[\dfrac{{{x}^{2}}}{{{2}^{2}}}+\dfrac{{{y}^{2}}}{{{5}^{2}}}=1\]

Here, the denominator of \[\dfrac{{{y}^{2}}}{25}\] is greater than the denominator of \[\dfrac{{{x}^{2}}}{4}\]. 

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis. 

On comparing the given equation with \[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1\], we’ll get \[b=2\] and \[a=5\] 

\[\begin{align}  & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\  & =\sqrt{25-4} \\  & =\sqrt{21} \\ \end{align}\]

Therefore, 

The coordinates of the foci are \[\left( 0,\sqrt{21} \right)\] and \[\left( 0,-\sqrt{21} \right)\]

The coordinates of the vertices are \[(0,5)\] and \[(0,-5)\]

Length of major axis \[=2a=10\]

Length of minor axis \[=2b=4\]

Eccentricity, \[e=\dfrac{c}{a}=\dfrac{\sqrt{21}}{5}\]

Length of latus rectum \[=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 4}{5}=\dfrac{8}{5}\]

3. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \[\dfrac{{{x}^{2}}}{16}+\dfrac{{{y}^{2}}}{9}=1\]

Ans: The given equation is \[\dfrac{{{x}^{2}}}{16}+\dfrac{{{y}^{2}}}{9}=1\] or  \[\dfrac{{{x}^{2}}}{{{4}^{2}}}+\dfrac{{{y}^{2}}}{{{3}^{2}}}=1\]

Here, the denominator of \[\dfrac{{{x}^{2}}}{16}\] is greater than the denominator of \[\dfrac{{{y}^{2}}}{9}\]. 

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis. 

On comparing the given equation with \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\], we’ll get \[a=4\] and \[b=3\]

\[\begin{align}  & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\  & =\sqrt{16-9} \\  & =\sqrt{7} \\ \end{align}\]

Therefore, 

The coordinates of the foci are \[\left( \pm \sqrt{7},0 \right)\]

The coordinates of the vertices are \[(\pm 4,0)\] 

Length of major axis \[=2a=8\]

Length of minor axis \[=2b=6\]

Eccentricity, \[e=\dfrac{c}{a}=\dfrac{\sqrt{7}}{4}\]

Length of latus rectum \[=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 9}{4}=\dfrac{9}{2}\]

4. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \[\dfrac{{{x}^{2}}}{25}+\dfrac{{{y}^{2}}}{100}=1\]

Ans: The given equation is \[\dfrac{{{x}^{2}}}{25}+\dfrac{{{y}^{2}}}{100}=1\] or  \[\dfrac{{{x}^{2}}}{{{5}^{2}}}+\dfrac{{{y}^{2}}}{{{10}^{2}}}=1\]

Here, the denominator of \[\dfrac{{{y}^{2}}}{100}\] is greater than the denominator of \[\dfrac{{{x}^{2}}}{25}\]. 

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis. 

On comparing the given equation with \[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1\], we’ll get \[b=5\] and \[a=10\]

\[\begin{align}  & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\  & =\sqrt{100-25} \\  & =\sqrt{75} \\  & =5\sqrt{3} \\ \end{align}\]

Therefore, 

The coordinates of the foci are \[\left( 0,\pm 5\sqrt{3} \right)\] 

The coordinates of the vertices are \[(0,\pm 10)\] 

Length of major axis \[=2a=20\]

Length of minor axis \[=2b=10\]

Eccentricity, \[e=\dfrac{c}{a}=\dfrac{5\sqrt{3}}{10}=\dfrac{\sqrt{3}}{2}\]

Length of latus rectum \[=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 25}{10}=5\]

5. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \[\dfrac{{{x}^{2}}}{49}+\dfrac{{{y}^{2}}}{36}=1\]

Ans: The given equation is \[\dfrac{{{x}^{2}}}{49}+\dfrac{{{y}^{2}}}{36}=1\] or  \[\dfrac{{{x}^{2}}}{{{7}^{2}}}+\dfrac{{{y}^{2}}}{{{6}^{2}}}=1\]

Here, the denominator of \[\dfrac{{{x}^{2}}}{49}\] is greater than the denominator of \[\dfrac{{{y}^{2}}}{36}\]. 

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis. 

On comparing the given equation with \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\], we’ll get \[a=7\]and \[b=6\]

\[\begin{align}  & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\  & =\sqrt{49-36} \\  & =\sqrt{13} \\ \end{align}\]

Therefore, 

The coordinates of the foci are \[\left( \pm \sqrt{13},0 \right)\] 

The coordinates of the vertices are \[(\pm 7,0)\] 

Length of major axis \[=2a=14\]

Length of minor axis \[=2b=12\]

Eccentricity, \[e=\dfrac{c}{a}=\dfrac{\sqrt{13}}{7}\]

Length of latus rectum \[=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 36}{7}=\dfrac{72}{7}\]

6. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \[\dfrac{{{x}^{2}}}{100}+\dfrac{{{y}^{2}}}{400}=1\]

Ans: The given equation is \[\dfrac{{{x}^{2}}}{100}+\dfrac{{{y}^{2}}}{400}=1\] or  \[\dfrac{{{x}^{2}}}{{{10}^{2}}}+\dfrac{{{y}^{2}}}{{{20}^{2}}}=1\]

Here, the denominator of \[\dfrac{{{y}^{2}}}{400}\] is greater than the denominator of \[\dfrac{{{x}^{2}}}{100}\]. 

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis. 

On comparing the given equation with \[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1\], we’ll get \[b=10\] and \[a=20\]

\[\begin{align}   & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\  & =\sqrt{400-100} \\  & =\sqrt{300} \\  & =10\sqrt{3} \\ \end{align}\]

Therefore, 

The coordinates of the foci are \[\left( 0,\pm 10\sqrt{3} \right)\] 

The coordinates of the vertices are \[(0,\pm 20)\] 

Length of major axis \[=2a=40\]

Length of minor axis \[=2b=20\]

Eccentricity, \[e=\dfrac{c}{a}=\dfrac{10\sqrt{3}}{20}=\dfrac{\sqrt{3}}{2}\]

Length of latus rectum \[=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 100}{20}=10\]

7. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \[36{{x}^{2}}+4{{y}^{2}}=144\]

Ans: The given equation is \[36{{x}^{2}}+4{{y}^{2}}=144\]. 

It can be written as 

\[36{{x}^{2}}+4{{y}^{2}}=144\]

Or, \[\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{36}=1\] 

Or, \[\dfrac{{{x}^{2}}}{{{2}^{2}}}+\dfrac{{{y}^{2}}}{{{6}^{2}}}=1\] ………(1)

Here, the denominator of \[\dfrac{{{y}^{2}}}{{{6}^{2}}}\]is greater than the denominator of \[\dfrac{{{x}^{2}}}{{{2}^{2}}}\]. 

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis. 

On comparing the given equation with \[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1\], we’ll get \[b=2\] and \[a=6\]

\[\begin{align}  & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\  & =\sqrt{36-4} \\  & =\sqrt{32} \\  & =4\sqrt{2} \\ \end{align}\]

Therefore, 

The coordinates of the foci are \[\left( 0,\pm 4\sqrt{2} \right)\] 

The coordinates of the vertices are \[(0,\pm 6)\] 

Length of major axis \[=2a=12\]

Length of minor axis \[=2b=4\]

Eccentricity, \[e=\dfrac{c}{a}=\dfrac{4\sqrt{2}}{6}=\dfrac{2\sqrt{2}}{3}\]

Length of latus rectum \[=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 4}{6}=\dfrac{4}{3}\]

8. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \[16{{x}^{2}}+{{y}^{2}}=16\]

Ans: The given equation is \[16{{x}^{2}}+{{y}^{2}}=16\]. 

It can be written as 

\[16{{x}^{2}}+{{y}^{2}}=16\]

Or, \[\dfrac{{{x}^{2}}}{1}+\dfrac{{{y}^{2}}}{16}=1\] 

Or, \[\dfrac{{{x}^{2}}}{{{1}^{2}}}+\dfrac{{{y}^{2}}}{{{4}^{2}}}=1\] ………(1)

Here, the denominator of \[\dfrac{{{y}^{2}}}{{{4}^{2}}}\]is greater than the denominator of \[\dfrac{{{x}^{2}}}{{{1}^{2}}}\]. 

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis. 

On comparing the given equation with \[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1\], we’ll get \[b=1\] and \[a=4\]

\[\begin{align}  & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\  & =\sqrt{16-1} \\  & =\sqrt{15} \\ \end{align}\]

Therefore, 

The coordinates of the foci are \[\left( 0,\pm \sqrt{15} \right)\] 

The coordinates of the vertices are \[(0,\pm 4)\] 

Length of major axis \[=2a=8\]

Length of minor axis \[=2b=2\]

Eccentricity, \[e=\dfrac{c}{a}=\dfrac{\sqrt{15}}{6}\]

Length of latus rectum \[=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 1}{4}=\dfrac{1}{2}\]

9. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \[4{{x}^{2}}+9{{y}^{2}}=36\]

Ans: The given equation is \[4{{x}^{2}}+9{{y}^{2}}=36\]. 

It can be written as 

\[4{{x}^{2}}+9{{y}^{2}}=36\]

Or, \[\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{4}=1\] 

Or, \[\dfrac{{{x}^{2}}}{{{3}^{2}}}+\dfrac{{{y}^{2}}}{{{2}^{2}}}=1\] ………(1)

Here, the denominator of \[\dfrac{{{x}^{2}}}{{{3}^{2}}}\]is greater than the denominator of \[\dfrac{{{y}^{2}}}{{{2}^{2}}}\]. 

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis. 

On comparing the given equation with \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\], we’ll get \[a=3\] and \[b=2\]

\[\begin{align}  & \therefore c=\sqrt{{{a}^{2}}-b{}^{2}} \\  & =\sqrt{9-4} \\  & =\sqrt{5} \\ \end{align}\]

Therefore, 

The coordinates of the foci are \[\left( \pm \sqrt{5},0 \right)\] 

The coordinates of the vertices are \[(\pm 3,0)\] 

Length of major axis \[=2a=6\]

Length of minor axis \[=2b=4\]

Eccentricity, \[e=\dfrac{c}{a}=\dfrac{\sqrt{5}}{3}\]

Length of latus rectum \[=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 4}{3}=\dfrac{8}{3}\]

10. Find the equation for the ellipse that satisfies the given conditions: Vertices \[(\pm 5,0),\]foci \[(\pm 4,0)\]. 

Ans: Vertices \[(\pm 5,0),\] foci \[(\pm 4,0)\] 

Here, the vertices are on the x-axis. 

Therefore, the equation of the ellipse will be of the form \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\], where a is the semi-major axis. 

Accordingly, \[a=5\] and \[c=4\]. 

It is known that \[{{a}^{2}}={{b}^{2}}+{{c}^{2}}\]. 

\[\begin{align}  & \therefore {{5}^{2}}={{b}^{2}}+{{4}^{2}} \\  & \Rightarrow 25={{b}^{2}}+16 \\  & \Rightarrow b=\sqrt{9}=3 \\ \end{align}\]

Thus, the equation of the ellipse is \[\dfrac{{{x}^{2}}}{{{5}^{2}}}+\dfrac{{{y}^{2}}}{{{3}^{2}}}=1\] or \[\dfrac{{{x}^{2}}}{25}+\dfrac{{{y}^{2}}}{9}=1\] 

11. Find the equation for the ellipse that satisfies the given conditions: Vertices \[(0,\pm 13),\] foci \[(0,\pm 5)\]

Ans:  Vertices \[(0,\pm 13),\] foci \[(0,\pm 5)\] 

Here, the vertices are on the y-axis. 

Therefore, the equation of the ellipse will be of the form \[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1\], where a is the semi-major axis. 

Accordingly, \[a=13\] and \[c=5\]. 

It is known that \[{{a}^{2}}={{b}^{2}}+{{c}^{2}}\]. 

\[\begin{align}  & \therefore {{13}^{2}}={{b}^{2}}+{{5}^{2}} \\  & \Rightarrow 169={{b}^{2}}+25 \\  & \Rightarrow b=\sqrt{144}=12 \\ \end{align}\]

Thus, the equation of the ellipse is \[\dfrac{{{x}^{2}}}{{{12}^{2}}}+\dfrac{{{y}^{2}}}{{{13}^{2}}}=1\] or \[\dfrac{{{x}^{2}}}{144}+\dfrac{{{y}^{2}}}{169}=1\]

12. Find the equation for the ellipse that satisfies the given conditions: Vertices \[(\pm 6,0),\] foci \[(\pm 4,0)\]

Ans: Vertices \[(\pm 6,0),\] foci \[(\pm 4,0)\] 

Here, the vertices are on the x-axis. 

Therefore, the equation of the ellipse will be of the form \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\], where a is the semi-major axis. 

Accordingly, \[a=6\] and \[c=4\]. 

It is known that \[{{a}^{2}}={{b}^{2}}+{{c}^{2}}\]. 

\[\begin{align}  & \therefore {{6}^{2}}={{b}^{2}}+{{4}^{2}} \\  & \Rightarrow 36={{b}^{2}}+16 \\  & \Rightarrow b=\sqrt{20} \\ \end{align}\]

Thus, the equation of the ellipse is \[\dfrac{{{x}^{2}}}{{{6}^{2}}}+\dfrac{{{y}^{2}}}{{{\left( \sqrt{20} \right)}^{2}}}=1\] or \[\dfrac{{{x}^{2}}}{36}+\dfrac{{{y}^{2}}}{20}=1\] 

13. Find the equation for the ellipse that satisfies the given conditions: Ends of major axis \[(\pm 3,0),\]ends of minor axis \[(0,\pm 2)\] 

Ans: Ends of major axis \[(\pm 3,0),\] ends of minor axis \[(0,\pm 2)\] 

Here, the major axis is along the x-axis. 

Therefore, the equation of the ellipse will be of the form \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\], where a is the semimajor axis. 

Accordingly, \[a=3\] and \[b=2\]. 

Thus, the equation of the ellipse is \[\dfrac{{{x}^{2}}}{{{3}^{2}}}+\dfrac{{{y}^{2}}}{{{2}^{2}}}=1\]or \[\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{4}=1\]

14. Find the equation for the ellipse that satisfies the given conditions: Ends of major axis \[(0,\pm \sqrt{5}),\] ends of minor axis \[(\pm 1,0)\]

Ans: Ends of major axis \[(0,\pm \sqrt{5}),\] ends of minor axis \[(\pm 1,0)\]

Here, the major axis is along the y-axis. 

Therefore, the equation of the ellipse will be of the form \[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1\], where a is the semimajor axis. 

Accordingly, \[a=\sqrt{5}\] and \[b=1\]. 

Thus, the equation of the ellipse is \[\dfrac{{{x}^{2}}}{{{1}^{2}}}+\dfrac{{{y}^{2}}}{{{\left( \sqrt{5} \right)}^{2}}}=1\] or \[\dfrac{{{x}^{2}}}{1}+\dfrac{{{y}^{2}}}{5}=1\]

15. Find the equation for the ellipse that satisfies the given conditions: Length of major axis \[26\], foci \[(\pm 5,0)\] 

Ans: Length of major axis = \[26\]; foci = \[(\pm 5,0)\]

Since the foci are on the x-axis, the major axis is along the x-axis. 

Therefore, the equation of the ellipse will be of the form \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\], where a is the semimajor axis. 

Accordingly, \[2a=26\Rightarrow a=13\] and \[c=5\]. 

It is known that \[{{a}^{2}}={{b}^{2}}+{{c}^{2}}\].

\[\begin{align}  & \therefore {{13}^{2}}={{b}^{2}}+{{5}^{2}} \\  & \Rightarrow 169={{b}^{2}}+25 \\  & \Rightarrow b=\sqrt{144}=12 \\ \end{align}\]

Thus, the equation of the ellipse is \[\dfrac{{{x}^{2}}}{{{13}^{2}}}+\dfrac{{{y}^{2}}}{{{12}^{2}}}=1\]or \[\dfrac{{{x}^{2}}}{169}+\dfrac{{{y}^{2}}}{144}=1\]

16. Find the equation for the ellipse that satisfies the given conditions: Length of minor axis \[16\], foci \[(0,\pm 6)\] 

Ans: Length of minor axis = \[16\], foci = \[(0,\pm 6)\]

Since the foci are on the y-axis, the major axis is along the y-axis. 

Therefore, the equation of the ellipse will be of the form \[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1\], where a is the semimajor axis. 

Accordingly, \[2b=16\Rightarrow b=8\] and \[c=6\]. 

It is known that \[{{a}^{2}}={{b}^{2}}+{{c}^{2}}\].

\[\begin{align}  & \therefore {{a}^{2}}={{8}^{2}}+{{6}^{2}} \\  & \Rightarrow {{a}^{2}}=64+36 \\  & \Rightarrow a=\sqrt{100}=10 \\ \end{align}\]

Thus, the equation of the ellipse is \[\dfrac{{{x}^{2}}}{{{8}^{2}}}+\dfrac{{{y}^{2}}}{{{10}^{2}}}=1\] or \[\dfrac{{{x}^{2}}}{64}+\dfrac{{{y}^{2}}}{100}=1\]

17. Find the equation for the ellipse that satisfies the given conditions: Foci \[(\pm 3,0),\] \[a=4\]

Ans: Foci \[(\pm 3,0),\]\[a=4\]

Since the foci are on the x-axis, the major axis is along the x-axis. 

Therefore, the equation of the ellipse will be of the form \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\], where a is the semimajor axis. 

Accordingly, \[c=3\] and \[a=4\]. 

It is known that \[{{a}^{2}}={{b}^{2}}+{{c}^{2}}\].

\[\begin{align}  & \therefore {{4}^{2}}={{b}^{2}}+{{3}^{2}} \\  & \Rightarrow 16={{b}^{2}}+9 \\  & \Rightarrow {{b}^{2}}=16-9=7 \\ \end{align}\]

Thus, the equation of the ellipse is \[\dfrac{{{x}^{2}}}{16}+\dfrac{{{y}^{2}}}{7}=1\]

18. Find the equation for the ellipse that satisfies the given conditions: \[b=3\], \[c=4\], centre at the origin; foci on the x axis. 

Ans: It is given that \[b=3\], \[c=4\], centre at the origin; foci on the x axis. 

Since the foci are on the x-axis, the major axis is along the x-axis. 

Therefore, the equation of the ellipse will be of the form \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\], where a is the semimajor axis. 

Accordingly, \[b=3\], \[c=4\]. 

It is known that \[{{a}^{2}}={{b}^{2}}+{{c}^{2}}\]. 

\[\begin{align}  & \therefore {{a}^{2}}={{3}^{2}}+{{4}^{2}} \\  & \Rightarrow {{a}^{2}}=9+16 \\  & \Rightarrow {{a}^{2}}=25 \\  & \Rightarrow a=\sqrt{25}=5 \\ \end{align}\]

Thus, the equation of the ellipse is \[\dfrac{{{x}^{2}}}{{{5}^{2}}}+\dfrac{{{y}^{2}}}{{{3}^{2}}}=1\] or \[\dfrac{{{x}^{2}}}{25}+\dfrac{{{y}^{2}}}{9}=1\]

19. Find the equation for the ellipse that satisfies the given conditions: Centre at \[(0,0)\], major axis on the y-axis and passes through the points \[(3,2)\] and \[(1,6)\].

Ans: Since the centre is at \[(0,0)\] and the major axis is on the y-axis, the equation of the ellipse will be of the form 

\[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1\] …..(1)

Where, a is the semi-major axis 

The ellipse passes through points \[(3,2)\] and \[(1,6)\]. Hence, 

\[\dfrac{9}{{{b}^{2}}}+\dfrac{4}{{{a}^{2}}}=1\] …..(2)

\[\dfrac{1}{{{b}^{2}}}+\dfrac{36}{{{a}^{2}}}=1\] …..(3)

On solving equations (2) and (3), we’ll get

 \[{{b}^{2}}=10\] and \[{{a}^{2}}=40\]. 

Thus, the equation of the ellipse is \[\dfrac{{{x}^{2}}}{10}+\dfrac{{{y}^{2}}}{40}=1\] or \[4{{x}^{2}}+{{y}^{2}}=40\]. 

20. Find the equation for the ellipse that satisfies the given conditions: Major axis on the x-axis and passes through the points \[(4,3)\] and \[(6,2)\]. 

Ans: Since the major axis is on the x-axis, the equation of the ellipse will be of the form 

\[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\] …..(1)

Where, a is the semi-major axis 

The ellipse passes through points \[(4,3)\] and \[(6,2)\]. Hence, 

\[\dfrac{16}{{{a}^{2}}}+\dfrac{9}{{{b}^{2}}}=1\] …..(2)

\[\dfrac{36}{{{a}^{2}}}+\dfrac{4}{{{b}^{2}}}=1\] …..(3)

On solving equations (2) and (3), we’ll get 

\[{{a}^{2}}=52\] and \[{{b}^{2}}=13\]. 

Thus, the equation of the ellipse is \[\dfrac{{{x}^{2}}}{52}+\dfrac{{{y}^{2}}}{13}=1\] or \[{{x}^{2}}+4{{y}^{2}}=52\]. 

Exercise 11.4

1. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola \[\dfrac{{{x}^{2}}}{16}-\dfrac{{{y}^{2}}}{9}=1\]

Ans: The given equation is \[\dfrac{{{x}^{2}}}{16}-\dfrac{{{y}^{2}}}{9}=1\] or \[\dfrac{{{x}^{2}}}{{{4}^{2}}}-\dfrac{{{y}^{2}}}{{{3}^{2}}}=1\]

On comparing this equation with the standard equation of hyperbola i.e., \[\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\], we’ll get \[a=4\] and \[b=3\]. 

We know that \[{{c}^{2}}={{a}^{2}}+{{b}^{2}}\]. 

\[\begin{align}  & \therefore {{c}^{2}}={{4}^{2}}+{{3}^{2}} \\  & \Rightarrow {{c}^{2}}=16+9 \\  & \Rightarrow {{c}^{2}}=25 \\  & \Rightarrow c=\sqrt{25}=5 \\ \end{align}\]

Therefore, 

The coordinates of the foci are \[\left( \pm 5,0 \right)\] 

The coordinates of the vertices are \[(\pm 4,0)\] 

Eccentricity, \[e=\dfrac{c}{a}=\dfrac{5}{4}\]

Length of latus rectum \[=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 9}{4}=\dfrac{9}{2}\]

2. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola \[\dfrac{{{y}^{2}}}{9}-\dfrac{{{x}^{2}}}{27}=1\]

Ans: The given equation is \[\dfrac{{{y}^{2}}}{9}-\dfrac{{{x}^{2}}}{27}=1\] or \[\dfrac{{{y}^{2}}}{{{3}^{2}}}-\dfrac{{{x}^{2}}}{{{\left( \sqrt{27} \right)}^{2}}}=1\]

On comparing this equation with the standard equation of hyperbola i.e., \[\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1\], we’ll get \[a=3\] and \[b=\sqrt{27}\]. 

We know that \[{{c}^{2}}={{a}^{2}}+{{b}^{2}}\]. 

\[\begin{align}  & \therefore {{c}^{2}}={{3}^{2}}+{{\left( \sqrt{27} \right)}^{2}} \\  & \Rightarrow {{c}^{2}}=9+27 \\ & \Rightarrow {{c}^{2}}=36 \\  & \Rightarrow c=\sqrt{36}=6 \\ \end{align}\]

Therefore, 

The coordinates of the foci are \[\left( 0,\pm 6 \right)\] 

The coordinates of the vertices are \[(0,\pm 3)\] 

Eccentricity, \[e=\dfrac{c}{a}=\dfrac{6}{3}=2\]

Length of latus rectum \[=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 27}{3}=18\]

3. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola \[9{{y}^{2}}-4{{x}^{2}}=36\]

Ans: The given equation is \[9{{y}^{2}}-4{{x}^{2}}=36\]. 

It can be written as 

\[9{{y}^{2}}-4{{x}^{2}}=36\]

Or, \[\dfrac{{{y}^{2}}}{4}-\dfrac{{{x}^{2}}}{9}=1\] 

Or, \[\dfrac{{{y}^{2}}}{{{2}^{2}}}-\dfrac{{{x}^{2}}}{{{3}^{2}}}=1\] ………(1)

On comparing equation (1) with the standard equation of hyperbola i.e., \[\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1\], we’ll get \[a=2\] and \[b=3\].

We know that \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]. 

\[\begin{align}  & \therefore {{c}^{2}}=4+9 \\  & \Rightarrow {{c}^{2}}=13 \\  & \Rightarrow c=\sqrt{13} \\ \end{align}\]

Therefore, 

The coordinates of the foci are \[\left( 0,\pm \sqrt{13} \right)\] 

The coordinates of the vertices are \[(0,\pm 2)\] 

Eccentricity, \[e=\dfrac{c}{a}=\dfrac{\sqrt{13}}{2}\]

Length of latus rectum \[=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 9}{2}=9\]

4. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola \[16{{x}^{2}}-9{{y}^{2}}=576\]

Ans: The given equation is \[16{{x}^{2}}-9{{y}^{2}}=576\]. 

It can be written as 

\[16{{x}^{2}}-9{{y}^{2}}=576\]

Or, \[\dfrac{{{x}^{2}}}{36}-\dfrac{{{y}^{2}}}{64}=1\] 

Or, \[\dfrac{{{x}^{2}}}{{{6}^{2}}}-\dfrac{{{y}^{2}}}{{{8}^{2}}}=1\] ………(1)

On comparing equation (1) with the standard equation of hyperbola i.e., \[\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\], we’ll get \[a=6\] and \[b=8\].

We know that \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]. 

\[\begin{align}  & \therefore {{c}^{2}}=36+64 \\  & \Rightarrow {{c}^{2}}=100 \\  & \Rightarrow c=\sqrt{100} \\  & \Rightarrow c=10 \\ \end{align}\]

Therefore, 

The coordinates of the foci are \[\left( \pm \sqrt{10},0 \right)\] 

The coordinates of the vertices are \[(\pm 6,0)\] 

Eccentricity, \[e=\dfrac{c}{a}=\dfrac{10}{6}=\dfrac{5}{3}\]

Length of latus rectum \[=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 64}{6}=\dfrac{64}{3}\]

5. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola \[5{{y}^{2}}-9{{x}^{2}}=36\]

Ans: The given equation is \[5{{y}^{2}}-9{{x}^{2}}=36\]. 

 \[\Rightarrow \dfrac{{{y}^{2}}}{\dfrac{36}{5}}-\dfrac{{{x}^{2}}}{4}=1\] 

 \[\dfrac{{{y}^{2}}}{\left( \dfrac{6}{\sqrt{5}} \right)^2}-\dfrac{{{x}^{2}}}{{{2}^{2}}}=1\] ………(1)

On comparing equation (1) with the standard equation of hyperbola i.e., \[\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1\], we’ll get \[a=\dfrac{6}{\sqrt{5}}\] and \[b=2\].

We know that \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]. 

\[\begin{align}  & \therefore {{c}^{2}}=\dfrac{36}{5}+4 \\  & \Rightarrow {{c}^{2}}=\dfrac{56}{5} \\  & \Rightarrow c=\sqrt{\dfrac{56}{5}} \\  & \Rightarrow c=\dfrac{2\sqrt{14}}{\sqrt{5}} \\ \end{align}\]

Therefore, 

The coordinates of the foci are \[\left( 0,\pm \dfrac{2\sqrt{14}}{\sqrt{5}} \right)\] 

The coordinates of the vertices are \[\left( 0,\pm \dfrac{6}{\sqrt{5}} \right)\] 

Eccentricity, \[e=\dfrac{c}{a}=\dfrac{\left( \dfrac{2\sqrt{14}}{\sqrt{5}} \right)}{\left( \dfrac{6}{\sqrt{5}} \right)}=\dfrac{\sqrt{14}}{3}\]

Length of latus rectum \[=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 4}{\left( \dfrac{6}{\sqrt{5}} \right)}=\dfrac{4\sqrt{5}}{3}\]

6. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola \[49{{y}^{2}}-16{{x}^{2}}=784\]

Ans: The given equation is \[49{{y}^{2}}-16{{x}^{2}}=784\]. 

It can be written as 

\[49{{y}^{2}}-16{{x}^{2}}=784\]

Or, \[\dfrac{{{y}^{2}}}{16}-\dfrac{{{x}^{2}}}{49}=1\] 

Or, \[\dfrac{{{y}^{2}}}{{{4}^{2}}}-\dfrac{{{x}^{2}}}{{{7}^{2}}}=1\] ………(1)

On comparing equation (1) with the standard equation of hyperbola i.e., \[\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1\], we’ll get \[a=4\] and \[b=7\].

We know that \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]. 

\[\begin{align}  & \therefore {{c}^{2}}=16+49 \\  & \Rightarrow {{c}^{2}}=65 \\  & \Rightarrow c=\sqrt{65} \\ \end{align}\]

Therefore, 

The coordinates of the foci are \[\left( 0,\pm \sqrt{65} \right)\] 

The coordinates of the vertices are \[(0,\pm 4)\] 

Eccentricity, \[e=\dfrac{c}{a}=\dfrac{\sqrt{65}}{4}\]

Length of latus rectum \[=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 49}{4}=\dfrac{49}{2}\]

7. Find the equation of the hyperbola satisfying the give conditions: Vertices \[(\pm 2,0)\], foci \[(\pm 3,0)\] 

Ans: Vertices \[(\pm 2,0)\], foci \[(\pm 3,0)\]

Here, the vertices are on the x-axis. 

Therefore, the equation of the hyperbola is of the form \[\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]

Since the vertices are \[(\pm 2,0)\], \[a=2\]. 

Since the foci are \[(\pm 3,0)\], \[c=3\].

We know that \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]. 

\[\begin{align}  & \therefore {{2}^{2}}+{{b}^{2}}={{3}^{2}} \\  & \Rightarrow {{b}^{2}}=9-4 \\  & \Rightarrow {{b}^{2}}=5 \\ \end{align}\]

Thus, the equation of the hyperbola is \[\dfrac{{{x}^{2}}}{4}-\dfrac{{{y}^{2}}}{5}=1\]

8. Find the equation of the hyperbola satisfying the give conditions: Vertices \[(0,\pm 5)\], foci \[(0,\pm 8)\]

Ans: Vertices \[(0,\pm 5)\], foci \[(0,\pm 8)\]

Here, the vertices are on the y-axis. 

Therefore, the equation of the hyperbola is of the form \[\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1\]

Since the vertices are \[(0,\pm 5)\], \[a=5\]. 

Since the foci are \[(0,\pm 8)\], \[c=8\].

We know that \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]. 

\[\begin{align}  & \therefore {{5}^{2}}+{{b}^{2}}={{8}^{2}} \\  & \Rightarrow {{b}^{2}}=64-25 \\  & \Rightarrow {{b}^{2}}=39 \\ \end{align}\]

Thus, the equation of the hyperbola is \[\dfrac{{{y}^{2}}}{25}-\dfrac{{{x}^{2}}}{39}=1\]

9. Find the equation of the hyperbola satisfying the give conditions: Vertices \[(0,\pm 3)\], foci \[(0,\pm 5)\] 

Ans: Vertices \[(0,\pm 3)\], foci \[(0,\pm 5)\]

Here, the vertices are on the y-axis. 

Therefore, the equation of the hyperbola is of the form \[\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1\]

Since the vertices are \[(0,\pm 3)\], \[a=3\]. 

Since the foci are \[(0,\pm 5)\], \[c=5\].

We know that \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]. 

\[\begin{align}   & \therefore {{3}^{2}}+{{b}^{2}}=25 \\  & \Rightarrow {{b}^{2}}=25-9 \\  & \Rightarrow {{b}^{2}}=16 \\ \end{align}\]

Thus, the equation of the hyperbola is \[\dfrac{{{y}^{2}}}{9}-\dfrac{{{x}^{2}}}{16}=1\]

10. Find the equation of the hyperbola satisfying the give conditions: Foci \[(\pm 5,0)\], the transverse axis is of length \[8\]. 

Ans: Foci \[(\pm 5,0)\], the transverse axis is of length \[8\]. 

Here, the foci are on the x-axis. 

Therefore, the equation of the hyperbola is of the form \[\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]. 

Since the foci are \[(\pm 5,0)\], \[c=5\]. 

Since the length of the transverse axis is \[8\], \[2a=8\Rightarrow a=4\]. 

We know that \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]. 

\[\begin{align}  & \therefore {{4}^{2}}+{{b}^{2}}=25 \\  & \Rightarrow {{b}^{2}}=25-16 \\  & \Rightarrow {{b}^{2}}=9 \\ \end{align}\]

Thus, the equation of the hyperbola is \[\dfrac{{{x}^{2}}}{16}-\dfrac{{{y}^{2}}}{9}=1\]

11. Find the equation of the hyperbola satisfying the give conditions: Foci \[(0,\pm 13)\], the conjugate axis is of length \[24\]. 

Ans: Foci \[(0,\pm 13)\], the transverse axis is of length \[24\]. 

Here, the foci are on the y-axis. 

Therefore, the equation of the hyperbola is of the form \[\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1\]. 

Since the foci are \[(0,\pm 13)\], \[c=13\]. 

Since the length of the transverse axis is \[24\], \[2b=24\Rightarrow b=12\]. 

We know that \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]. 

\[\begin{align}  & \therefore {{a}^{2}}+{{12}^{2}}={{13}^{2}} \\  & \Rightarrow {{a}^{2}}=169-144 \\  & \Rightarrow {{a}^{2}}=25 \\ \end{align}\]

Thus, the equation of the hyperbola is \[\dfrac{{{y}^{2}}}{25}-\dfrac{{{x}^{2}}}{144}=1\]

12. Find the equation of the hyperbola satisfying the give conditions: Foci \[\left( \pm 3\sqrt{5},0 \right)\], the latus rectum is of length \[8\]. 

Ans: Foci \[\left( \pm 3\sqrt{5},0 \right)\], the latus rectum is of length \[8\]. 

Here, the foci are on the x-axis. 

Therefore, the equation of the hyperbola is of the form \[\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]

Since the foci are \[\left( \pm 3\sqrt{5},0 \right)\], \[c=\pm 3\sqrt{5}\]

Length of latus rectum = \[8\]

\[\begin{align}  & \Rightarrow \dfrac{2{{b}^{2}}}{a}=8 \\  & \Rightarrow {{b}^{2}}=4a \\ \end{align}\]

We know that \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]. 

\[\begin{align}  & \therefore {{a}^{2}}+4a=45 \\  & \Rightarrow {{a}^{2}}+4a-45=0 \\  & \Rightarrow {{a}^{2}}+9a-5a-45=0 \\  & \Rightarrow \left( a+9 \right)\left( a-5 \right)=0 \\  & \Rightarrow a=-9,5 \\ \end{align}\]

Since a is non-negative, \[a=5\]. 

\[\therefore {{b}^{2}}=4a=4\times 5=20\]

Thus, the equation of the hyperbola is \[\dfrac{{{x}^{2}}}{25}-\dfrac{{{y}^{2}}}{20}=1\]

13. Find the equation of the hyperbola satisfying the give conditions: Foci \[\left( \pm 4,0 \right)\], the latus rectum is of length \[12\]. 

Ans: Foci \[\left( \pm 4,0 \right)\], the latus rectum is of length \[12\]. 

Here, the foci are on the x-axis. 

Therefore, the equation of the hyperbola is of the form \[\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]

Since the foci are \[\left( \pm 4,0 \right)\], \[c=4\]. 

Length of latus rectum = \[12\]

\[\begin{align}  & \Rightarrow \dfrac{2{{b}^{2}}}{a}=12 \\  & \Rightarrow {{b}^{2}}=6a \\ \end{align}\]

We know that \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]. 

\[\begin{align}  & \therefore {{a}^{2}}+6a=16 \\ & \Rightarrow {{a}^{2}}+6a-16=0 \\  & \Rightarrow {{a}^{2}}+8a-2a-16=0 \\  & \Rightarrow \left( a+8 \right)\left( a-2 \right)=0 \\ & \Rightarrow a=-8,2 \\ \end{align}\]

Since a is non-negative, \[a=2\]. 

\[\therefore {{b}^{2}}=6a=6\times 2=12\]

Thus, the equation of the hyperbola is \[\dfrac{{{x}^{2}}}{4}-\dfrac{{{y}^{2}}}{12}=1\]

14. Find the equation of the hyperbola satisfying the give conditions: Vertices \[\left( \pm 7,0 \right)\], \[e=\dfrac{4}{3}\]

Ans: Vertices \[\left( \pm 7,0 \right)\], \[e=\dfrac{4}{3}\] 

Here, the vertices are on the x-axis. 

Therefore, the equation of the hyperbola is of the form \[\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]

Since the vertices are \[\left( \pm 7,0 \right)\], \[a=7\]. 

It is given that \[e=\dfrac{4}{3}\]

\[\therefore \dfrac{c}{a}=\dfrac{4}{3}\] \[\left[ e=\dfrac{c}{a} \right]\]

\[\begin{align}  & \Rightarrow \dfrac{c}{7}=\dfrac{4}{3} \\  & \Rightarrow c=\dfrac{28}{3} \\ \end{align}\]

We know that \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]. 

\[\begin{align}  & \therefore {{7}^{2}}+{{b}^{2}}={{\left( \dfrac{28}{3} \right)}^{2}} \\  & \Rightarrow {{b}^{2}}=\dfrac{784}{9}-49 \\  & \Rightarrow {{b}^{2}}=\dfrac{784-441}{9}=\dfrac{343}{9} \\ \end{align}\]

Thus, the equation of the hyperbola is \[\dfrac{{{x}^{2}}}{49}-\dfrac{{9{y}^{2}}}{343}=1\]

15. Find the equation of the hyperbola satisfying the given conditions: Foci \[\left( 0,\pm \sqrt{10} \right)\], passing through \[\left( 2,3 \right)\]

Ans: Foci \[\left( 0,\pm \sqrt{10} \right)\], passing through \[\left( 2,3 \right)\] 

Here, the foci are on the y-axis. 

Therefore, the equation of the hyperbola is of the form \[\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1\] .

Since the foci are \[\left( 0,\pm \sqrt{10} \right)\], \[c=\sqrt{10}\] . 

We know that \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]. 

\[\therefore {{a}^{2}}+{{b}^{2}}=10\]

\[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]. 

\[\Rightarrow {{b}^{2}}=10-{{a}^{2}}\] … (1) 

Since the hyperbola passes through point \[\left( 2,3 \right)\], 

\[\dfrac{9}{{{a}^{2}}}-\dfrac{4}{{{b}^{2}}}=1\] ….(2) 

From equations (1) and (2), we’ll get  

\[\begin{align}  & \dfrac{9}{{{a}^{2}}}-\dfrac{4}{{{\left( 10-a^{2}\right)}}}=1 \\  & \Rightarrow 9\left( 10-{{a}^{2}} \right)-4{{a}^{2}}={{a}^{2}}\left( 10-{{a}^{2}} \right) \\  & \Rightarrow 90-9{{a}^{2}}-4{{a}^{2}}=10{{a}^{2}}-{{a}^{4}} \\  & \Rightarrow {{a}^{4}}-10{{a}^{2}}-9{{a}^{2}}-4{{a}^{2}}+90=0 \\  & \Rightarrow {{a}^{4}}-23{{a}^{2}}+90=0 \\  & \Rightarrow {{a}^{4}}-18{{a}^{2}}-5{{a}^{2}}+90=0 \\  & \Rightarrow {{a}^{2}}({{a}^{2}}-18)-5({{a}^{2}}-18)=0 \\  & \Rightarrow ({{a}^{2}}-5)({{a}^{2}}-18)=0 \\  & \Rightarrow {{a}^{2}}=18,5 \\ \end{align}\]

In hyperbola, \[c>a\] i.e., \[{{c}^{2}}>{{a}^{2}}\]

\[\begin{align}  & \therefore a{}^{2}=5 \\  & \Rightarrow {{b}^{2}}=10-{{a}^{2}} \\  & \Rightarrow {{b}^{2}}=10-5=5 \\ \end{align}\]

Thus, the equation of the hyperbola is \[\dfrac{{{y}^{2}}}{5}-\dfrac{{{x}^{2}}}{5}=1\]

Miscellaneous Exercise

1. If a parabolic reflector is \[20\] cm in diameter and \[5\]cm deep, find the focus.

Ans: As we know that the origin of the coordinate plane is taken at the vertex of the parabolic reflector, where the axis of the reflector is along the positive x-axis.

The Diagrammatic representation is represented as follows:

(Image will be uploaded soon)

As we know that the equation of the parabola is of the form of \[{{y}^{2}}=4ax\] (as it is opening to the right)

Since, the parabola passes through point \[A \left( 5,10 \right),\]

\[\begin{align}  & \Rightarrow {{y}^{2}}=4ax \\  & \Rightarrow {{10}^{2}}=4\times a\times 5 \\  & \Rightarrow 100=20a \\  & \Rightarrow a=\dfrac{100}{20} \\  & \Rightarrow a=5 \\ \end{align}\]

The focus of the parabola is \[\left( a,0 \right)=\left( 5,0 \right),\] which is the mid–point of the diameter.

Hence, the focus of the reflector is at the midpoint of the diameter.

2. An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?

Ans: As we know that the origin of the coordinate plane is taken at the vertex of the arch in such a way that its axis is along the y-axis.

The diagrammatic representation will be as follows:

(Image will be uploaded soon)

The equation of the parabola is of the form \[{{x}^{2}}=-4ay\] (as it is opening downwards). 

Since the parabola passes through point \[\left( \dfrac{5}{2},-10 \right),\]

\[\begin{align}  & {{\left( \dfrac{5}{2} \right)}^{2}}=-4\times a\times (-10) \\  & \Rightarrow a=\dfrac{25}{4\times 4\times 10}=\dfrac{5}{32} \\ \end{align}\]

Therefore,

The arch is in the form of a parabola whose equation is \[{{x}^{2}}=-\dfrac{5}{8}y\]

When \[y=-2m,{{x}^{2}}=-\dfrac{5}{8}\times (-2)\] because it's 2m away downward from the vertex, so it will be negative.

\[\begin{align}  & \Rightarrow {{x}^{2}}=\dfrac{5}{4} \\  & \Rightarrow x=\sqrt{\dfrac{5}{4}}m \\ \end{align}\]

Width of parabola at 2m away from vertex will be $=2\times \sqrt{\dfrac{5}{4}}m$

$=2\times 1.118m$(approx.) 

$=2.23m$(approx.) 

Hence, when the arch is 2 m from the vertex of the parabola, its width is approximately 2.23m.

3. The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.

Ans: The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the positive y-axis. 

This can be diagrammatically represented as

(image will be uploaded soon)

Here, AB and OC are the longest and shortest wires, respectively, attached to the cable.

DF is the supporting wire attached to the roadway, 18m from the middle.

Here, AB\[=30m, OC=6m, and\;  BC=\dfrac{100}{2}=50m\]

The equation of the parabola is of the form \[{{x}^{2}}=4ay\] (as it is opening upwards).

The coordinates of point A are \[(50,30-6)=(50,24)\].

Since A\[(50,24)\] is a point on the parabola,

\[\begin{align}   & {{(50)}^{2}}=4a(24) \\  & \Rightarrow a=\dfrac{50\times 50}{4\times 24}=\dfrac{625}{24} \\ \end{align}\]

\[\therefore \]Equation of the parabola, \[{{x}^{2}}=4\times \dfrac{625}{24}\times y\] or \[6{{x}^{2}}=625y\]

The x-coordinate of point D is 18. 

Hence, at x = 18, 

\[\begin{align}  & 6{{(18)}^{2}}=625y \\  & \Rightarrow y=\dfrac{6\times 18\times 18}{625} \\  & \Rightarrow y=3.11(approx.) \\ \end{align}\]

\[\therefore \]DE = \[3.11\]m 

DF = DE + EF = \[3.11\] m + \[6\] m = \[9.11\] m 

Thus, the length of the supporting wire attached to the roadway 18 m from the middle is approximately \[9.11\] m.

4. An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

Ans: Since the height and width of the arc from the centre is 2 m and 8 m respectively, it is clear that the length of the major axis is 8 m, while the length of the semi-minor axis is 2 m. 

The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken along the x-axis. 

Hence, the semi-ellipse can be diagrammatically represented as

(Image will be uploaded soon)

The equation of the semi-ellipse will be of the form \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1,y\ge 0,\] where a is the semi major axis 

Accordingly,

\[\begin{align}  & 2a=8\Rightarrow a=4 \\  & b=2 \\ \end{align}\]

Therefore, the equation of the semi-ellipse is \[\dfrac{{{x}^{2}}}{16}+\dfrac{{{y}^{2}}}{4}=1,y\ge 0,\] ….(1)

Let A be a point on the major axis such that AB = 1.5 m. 

Draw AC ⊥ OB.

OA = (4 – 1.5) m = 2.5 m 

The x-coordinate of point C is 2.5. 

On substituting the value of x with 2.5 in equation (1), we’ll get 

\[\begin{align}  & \dfrac{{{\left( 2.5 \right)}^{2}}}{16}+\dfrac{{{y}^{2}}}{4}=1 \\  & \Rightarrow \dfrac{6.25}{16}+\dfrac{{{y}^{2}}}{4}=1 \\  & \Rightarrow {{y}^{2}}=4\left( 1-\dfrac{6.25}{16} \right) \\  & \Rightarrow {{y}^{2}}=4\left( \dfrac{9.75}{16} \right) \\  & \Rightarrow {{y}^{2}}=2.4375 \\  & \Rightarrow {y}=1.56(approx.) \\  & \therefore AC=1.56m \\ \end{align}\]

Thus, the height of the arch at a point 1.5 m from one end is approximately 1.56 m.

5. A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

Ans: Let AB be the rod making an angle θ with OX and P (x, y) be the point on it such that AP = 3 cm. 

Then, PB = AB – AP = (12 – 3) cm = 9 cm [AB = 12 cm] 

From P, draw PQ⊥OY and PR⊥OX.

(image will be uploaded soon)

In ΔPBQ, \[\cos \theta =\dfrac{PQ}{PB}=\dfrac{x}{9}\]

In ΔPRA, \[\sin \theta =\dfrac{PR}{PA}=\dfrac{y}{3}\]

Since, $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,$

$ {{\left( \dfrac{y}{3} \right)}^{2}}+{{\left( \dfrac{x}{9} \right)}^{2}}=1 $

Or,  \[{{\dfrac{x}{81}}^{2}}+{{\dfrac{y}{9}}^{2}}=1\]

Thus, the equation of the locus of point P on the rod is \[{{\dfrac{x}{81}}^{2}}+{{\dfrac{y}{9}}^{2}}=1\]

6. Find the area of the triangle formed by the lines joining the vertex of the parabola \[{{x}^{2}}=12y\] to the ends of its latus rectum.

Ans: The given parabola is \[{{x}^{2}}=12y\]. 

On comparing this equation with \[{{x}^{2}}=4ay\], we’ll get \[4a=12\Rightarrow a=3\]

\[\therefore \]The coordinates of foci are S (0, a) = S (0, 3) 

Let AB be the latus rectum of the given parabola. 

The given parabola can be roughly drawn as

(image will be uploaded soon)

At \[y=3,\] \[{{x}^{2}}=12(3)\Rightarrow {{x}^{2}}=36\Rightarrow x=\pm 6\] 

\[\therefore \]The coordinates of A are (–6, 3), while the coordinates of B are (6, 3). 

Therefore, the vertices of ΔOAB are O (0, 0), A (–6, 3), and B (6, 3). 

Area of ΔOAB \[=\dfrac{1}{2}\left| 0\left( 3-3 \right)+\left( -6 \right)\left( 3-0 \right)+6\left( 0-3 \right) \right|uni{{t}^{2}}\]

$ =\dfrac{1}{2}\left| \left( -6 \right)\left( 3 \right)+6\left( -3 \right) \right|uni{{t}^{2}} $

$  =\dfrac{1}{2}\left| -18-18 \right|uni{{t}^{2}} $

$  =\dfrac{1}{2}\left| -36 \right|uni{{t}^{2}} $

\[=\dfrac{1}{2}\times 36uni{{t}^{2}}\]

\[=18\] \[uni{{t}^{2}}\]

Thus, the required area of the triangle is \[18\] \[uni{{t}^{2}}\].

7. A man running a racecourse notes that the sum of the distances from the two flag posts form him is always 10 m and the distance between the flag posts is 8 m. find the equation of the posts traced by the man.

Ans: Let A and B be the positions of the two flag posts and P(x, y) be the position of the man. Accordingly, PA + PB = 10.

We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Therefore, the path described by the man is an ellipse where the length of the major axis is 10 m, while points A and B are the foci.

Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis along the x-axis, the ellipse can be diagrammatically represented as

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The equation of the ellipse will be of the form of \[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1,\] where a is the semi-major axis.

Accordingly, \[2a=10\Rightarrow a=5\]

Distance between the foci \[(2c)=8\Rightarrow c=4\]

On using the relation \[c=\sqrt{{{a}^{2}}-{{b}^{2}}},\] we’ll get

$ 4=\sqrt{25-{{b}^{2}}} $

$ \Rightarrow 16=25-{{b}^{2}} $ 

$  \Rightarrow {{b}^{2}}=25-16 $ 

$  \Rightarrow {{b}^{2}}=9 $

$  \Rightarrow b=3 $

Thus, the equation of the path traced by the man is \[\dfrac{{{x}^{2}}}{25}+\dfrac{{{y}^{2}}}{9}=1\]

8. An equilateral triangle is inscribed in the parabola \[{{y}^{2}}=4ax\], where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Ans: Let OAB be the equilateral triangle inscribed in parabola \[{{y}^{2}}=4ax\]

Let AB intersect the x-axis at point C.

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Let OC = k 

From the equation of the given parabola, we have \[{{y}^{2}}=4ak\Rightarrow y=\pm 2\sqrt{ak}\]

\[\therefore \]The respective coordinates of points A and B are \[\left( k,-2\sqrt{ak} \right)\] and \[\left( k,-2\sqrt{ak} \right)\]

AB = CA + CB = \[2\sqrt{ak}+2\sqrt{ak}=4\sqrt{ak}\]

Since OAB is an equilateral triangle, \[O{{A}^{2}}=A{{B}^{2}}.\]

$ \therefore {{k}^{2}}+{{\left( 2\sqrt{ak} \right)}^{2}}={{\left( 4\sqrt{ak} \right)}^{2}} $

$  \Rightarrow {{k}^{2}}+4ak=16ak $

$  \Rightarrow {{k}^{2}}=12ak $

$  \Rightarrow k=12a $

$ \therefore AB=4\sqrt{ak}=4\sqrt{a\times 12a} $ 

$ =4\sqrt{12{{a}^{2}}} $

$ =8\sqrt{3}a $ 

Thus, the side of the equilateral triangle inscribed in parabola \[{{y}^{2}}=4ax\] is \[8\sqrt{3}a\]

NCERT Solutions For Class 11 Math Chapter 11 Conic Section

NCERT Solutions for Class 11 Math Chapter 11 Conic Sections are provided here in the pdf format with an aim to help students for their Math exam as well as to score better marks. The expert Math teacher at Vedantu has developed these NCERT Solutions for Chapter 11 as per the latest syllabus issued by the CBSE. 

Chapter 11 which is included in Class 11 Math NCERT aggregates some of the important topics such as Introduction to Conic Sections, Sections of a Circle as well as Circle, Parabola, Hyperbola and Ellipse. So, in order to gain sufficient information on all these topics, students can rely on NCERT Solutions. Using these solutions, the students can practise different challenging questions of the chapter to enhance their question-solving skills.

Topics Covered In Class 11 Math Chapter 11

11.1: Introduction to Conic Section

11.2: Section of A Cone

11.2.1: Ellipse, Circle, Parabola, Hyperbola

11.2.2: Degenerate Conic Section

11.3: Circle

11.4: Parabola

11.4.1: Standard Equation of a Parabola

11.4.2: Latus Rectum

11.5: Ellipse

11.6: Hyperbola

11.6.1: Eccentricity

11.6.2: Standard Equation of a parabola

11.6.3: Latus Rectum

NCERT Solutions For Class 11 Maths Chapter 11 - Free PDF Download

While solving exercise questions, you may struggle with basic problems, which happens due to lack of research or information regarding specific sections of the chapter. Since Ch 11 Maths Class 11 includes interesting topics, grasping the underlying concepts becomes automatically easier.

It gets quite difficult at times to find a reliable study material as there are numerous options available all over the internet. Among them, Conic Section Class 11 NCERT Solution follow an easy approach so that students get to understand the basics while solving these problems. Class 11 Conic Section Solution PDF can be easily accessed from renowned educational sites like Vedantu without any charge, to get a better understanding of the fundamentals as well as tricky areas.

Different Questions Derived from Class 11 Conic Section

Conic Section Class 11 deals with different shapes, circular cones, etc. and the core concept of it has been discussed throughout the chapter. It also contains several sections where you get to learn more about the semi-minor axis, the distance between focus and the central point of an ellipse, lots of equations related to that, etc.

In order to get satisfactory results in this particular subject, you should be aware of each and every part. Conic Sections Class 11 NCERT Solutions have been presented by keeping all these things in mind so that grasping these concepts becomes easier.

Here are some questions to help you understand the basic requirements:

• In the first few questions, (1-5) students are required to solve the equations containing different radius and centre of a circle. In order to solve the equation accurately, you should be aware of the fundamentals of the related section.

• In the next set of questions, (6-9) different equations have been given, and students will have to find the radius and centre of these circles.

You will encounter some unique set of questions as well that eventually help during examinations.

We cover all exercises in the chapter given below:- 

• Exercise 11.1 - 15 Questions with Solutions.

• Exercise 11.2 - 12 Questions with Solutions.

• Exercise 11.3 - 20 Questions with Solutions.

• Exercise 11.4 - 15 Questions with Solutions.

Benefits of NCERT Solutions Class 11 Maths Chapter 11

Class 11 Maths Ch 11 NCERT Solutions can be considered as useful resource material as it contains lots of strategies that help to achieve better results. Apart from that, students should also be prepared for uncommon or tricky questions before they face them in finals.

Since Conic Sections NCERT Solutions maintain CBSE guidelines, the formatting and patterns remain the same and students do not struggle to find relevant resources.

Some of the advantages also include:

• Fast-solving techniques to help students finish the paper on time.

• Accurate solutions crafted by subject matter experts that automatically bring excellent scores.

• Advanced methods that guide you properly if you are aiming to opt for Mathematics in your higher education.

• Conic Sections Class 11 Solutions also help you to get a hold of the core concepts as you go through.