Important Questions for CBSE Class 11 Maths Chapter 11 - Conic Sections

Very Short Answer Questions: (1 Marks)

1. Find the equation of a circle with centre (P,Q) & touching the y -  axis

(A) $\mathbf{{x^2} + {y^2} + 2Qy + {Q^2} = 0}$

(B)  $\mathbf{{x^2} + {y^2} - 2px - 2Qy + {Q^2} = 0}$

(C) $\mathbf{{x^2} + {y^2} - 2px + 2Qy + {Q^2} = 0}$

(D) None of these

Ans: ${x^2} + {y^2} - 2px - 2Qy + {Q^2} = 0$

2. Find the equations of the directrix & the axis of the parabola $\mathbf{\Rightarrow 3{x^2} = 8y}$

(A) $\mathbf{3y - 4 = 0,{\text{ }}x = 0}$

(B) $\mathbf{3x - 2 = 0,{\text{ }}X = 0}$

(C) $\mathbf{3y - 4x = 0}$

(D) None of these

Ans: $3y - 4 = 0,{\text{ }}x = 0$

3. Find the coordinates of the foci of the ellipse $ \mathbf{\Rightarrow {x^2} + 4{y^2} = 100}$

(A) $\mathbf{F( \pm 5\sqrt 3 ,0)}$

(B) $\mathbf{F( \pm 3\sqrt 5 ,0)}$

(C) $\mathbf{F( \pm 4\sqrt 5 ,0)}$

(D) None of these

Ans: $F( \pm 5\sqrt 3 ,0)$

4. Find the eccentricity of the hyperbola: $\mathbf{3{x^2} - 2{y^2} = 6}$

1. $\mathbf{e = \sqrt {\dfrac{5}{2}}} $

2. $\mathbf{e = \dfrac{{\sqrt 5 }}{2}}$

3. $\mathbf{e = \dfrac{{\sqrt 2 }}{5}}$

4. None of these

Ans: $e = \sqrt {\dfrac{5}{2}} $

5. Find the equation of a circle with centre $\mathbf{(b,a)}$ & touching $\mathbf{x - }$ axis?

1. $\mathbf{{x^2} + {y^2} - 2bx + 2ay + {b^2} = 0}$

2. $\mathbf{{x^2} + {y^2} + 2bx - 2ay + {b^2} = 0}$

3. $\mathbf{{x^2} + {y^2} - 2bx - 2ay + {b^2} = 0}$

4. None of these

Ans: ${x^2} + {y^2} - 2bx - 2ay + {b^2} = 0$

6. Find the length of the latus rectum of $\mathbf{2{x^2} + 3{y^2} = 18?}$

1. $\mathbf{2}$ units

2. $\mathbf{3}$ units

3. $\mathbf{4}$ units

4. None of these

Ans: $4$ units

7. Find the length of the latus rectum of the parabola $\mathbf{3{y^2} = 8x}$

1. $\mathbf{\dfrac{4}{3}}$ units

2. $\mathbf{\dfrac{8}{3}}$ units

3. $\mathbf{\dfrac{2}{3}}$ units

4. None of these

Ans: $\dfrac{8}{3}$ units

8. The equation $\mathbf{{x^2} + {y^2} - 12x + 8y - 72 = 0}$ represent a circle find its centre

1. \[\mathbf{( - 6, - 4)}\]

2. \[\mathbf{(6, - 4)}\]

3.  \[\mathbf{(6,4)}\]

4. \[\mathbf{( - 6,4)}\]

Ans: $(6, - 4)$

9. Find the equation of the parabola with focus $\mathbf{F(4,0)}$ & directrix $\mathbf{x =  - 4}$

1. $\mathbf{{y^2} = 32x}$

2. $\mathbf{{y^2} =  - 16x}$

3. $\mathbf{{y^2} = 8x}$

4. $\mathbf{{y^2} = 16x}$

Ans: ${y^2} = 16x$

10. Find the coordinates of the foci of $\mathbf{\dfrac{{{x^2}}}{8} + \dfrac{{{y^2}}}{4} = 1}$

1. $\mathbf{{F_1}(2,0)}$& $\mathbf{{F_2}( - 2,0)}$

2. $\mathbf{{F_1}( - 2,0)}$& $\mathbf{{F_2}(2,0)}$

3. $\mathbf{{F_1}( - 2,0)}$& $\mathbf{{F_2}( - 2,0)}$

4. None of these

Ans: ${F_1}( - 2,0)$& ${F_2}(2,0)$

11. Find the coordinates of the vertices of $\mathbf{{x^2} - {y^2} = 1}$

1. $\mathbf{A( - 1,0),{\text{ }}B( - 1,0)}$

2. $\mathbf{A( - 1,0),{\text{ }}B(1,0)}$

3. $ \mathbf{- A(1,0),{\text{ }}B( - 1,0)}$

4. None of these

Ans: $A( - 1,0),{\text{ }}B(1,0)$

13. Find the eccentricity of ellipse $\mathbf{4{x^2} + 9{y^2} = 1}$

1. $\mathbf{e = \dfrac{{\sqrt 5 }}{3}}$

2. $\mathbf{e = \dfrac{{ - \sqrt 5 }}{3}}$

3. $\mathbf{e = \dfrac{{\sqrt 3 }}{5}}$

4. $\mathbf{e = \dfrac{3}{{\sqrt 5 }}}$

Ans: $e = \dfrac{{\sqrt 5 }}{3}$

14. Find the length of the latus rectum of $\mathbf{9{x^2} + {y^2} = 36}$

1. $\mathbf{\dfrac{1}{3}}$ units

2. $\mathbf{\dfrac{1}{5}}$ units

3. $\mathbf{1\dfrac{1}{3}}$ units

4. $\mathbf{\dfrac{1}{6}}$ units

Ans: $1\dfrac{1}{3}$ units

15. Find the length of minor axis of $\mathbf{{x^2} + 4{y^2} = 100}$

1. $\mathbf{10}$ units

2. $\mathbf{12}$ units

3. $\mathbf{14}$ units

4. $\mathbf{8}$ units

Ans: $10$ units 

16. Find the centre of the circles $\mathbf{{x^2} + {(y - 1)^2} = 2}$

1. $\mathbf{(1,0)}$

2. $\mathbf{(0,1)}$ 

3. $\mathbf{(1,2)}$ 

4. None of these

Ans: $(0,1)$

17. Find the radius of circles $\mathbf{{x^2} + {(y - 1)^2} = 2}$

1. $\mathbf{\sqrt 2 }$

2. $\mathbf{2}$ 

3. $\mathbf{2\sqrt 2} $ 

4. None of these

Ans: $\sqrt 2 $

18. Find the length of latus rectum of $\mathbf{{x^2} =  - 22y}$

1. \[\mathbf{11}\]

2. $ \mathbf{- 22}$ 

3. $\mathbf{22}$ 

4. None of these

Ans: $22$

19. Find the length of latus rectum of $\mathbf{25{x^2} + 4{y^2} = 100}$

1. $\mathbf{\dfrac{3}{5}}$ units

2. $\mathbf{\dfrac{1}{5}}$ units 

3. $\mathbf{\dfrac{8}{5}}$ units 

4. None of these

Ans: $\dfrac{8}{5}$ units

Long Answer Questions: (4 Marks)

1. Show that the equation $\mathbf{{x^2} + {y^2} - 6x + 4y - 36 = 0}$ represents a circle, also find its centre & radius?

Ans: It is of the form ${x^2} + {y^2} + 2gx + 2Fy + c = 0$,

Where $2g =  - 6,{\text{ }}2f = 4$& $c =  - 36$

$\therefore g =  - 3,{\text{ }}f = 2$& $c =  - 36$

Thus, center of the circle is $( - g, - f) = (3, - 2)$

Radius of the circle is $\sqrt {{g^2} + {f^2} - c}  = \sqrt {9 + 4 + 36} $

$ = 7$ units

2. Find the equation of an ellipse whose foci are $\mathbf{( \pm 8,0)}$ & the eccentricity is $\mathbf{\dfrac{1}{4}}$ ?

Ans: Let the required equation of the ellipse be $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$, where ${a^2} > {b^2}$

Let the foci be $( \pm c,0),{\text{ }}c = 8$

And $e = \dfrac{c}{a}$

$ \Rightarrow a = \dfrac{c}{e}$

$ \Rightarrow a = \dfrac{8}{{\dfrac{1}{4}}}$

$ \Rightarrow a = 32$

As ${c^2} = {a^2} - {b^2}$

$ \Rightarrow {b^2} = {a^2} - {c^2}$

$ \Rightarrow {b^2} = 1024 - 64$

$ \Rightarrow {b^2} = 960$

$\therefore {a^2} = 1024$

& ${b^2} = 960$

Therefore the equation is $\dfrac{{{x^2}}}{{1024}} + \dfrac{{{y^2}}}{{960}} = 1$

3. Find the equation of an ellipse whose vertices are $\mathbf{(0, \pm 10)}$& $\mathbf{e = \dfrac{4}{5}}$

Ans: Let equation be $\dfrac{{{x^2}}}{{{b^2}}} + \dfrac{{{y^2}}}{{{a^2}}} = 1$

Vertices are $(0, \pm a),{\text{ }}a = 10$

Let ${c^2} = {a^2} - {b^2}$

So, $e = \dfrac{c}{a}$

$ \Rightarrow c = ae$

$ \Rightarrow c = 10 \times \dfrac{4}{5}$

$ \Rightarrow c = 8$

Now, ${c^2} = {a^2} - {b^2}$

$ \Rightarrow {b^2} = \left( {{a^2} - {c^2}} \right)$

$ \Rightarrow {b^2} = 100 - 64$

$ \Rightarrow {b^2} = 36$

So, ${a^2} = {(10)^2} = 100$ & ${b^2} = 36$

Therefore the equation is $\dfrac{{{x^2}}}{{36}} + \dfrac{{{y^2}}}{{100}} = 1$

4. Find the equation of hyperbola whose length of latus rectum is $\mathbf{36}$ & foci are $\mathbf{(0, \pm 12)}$

Ans: It is clear that $c = 12$.

Length of latus rectum $ = 36$

$ \Rightarrow \dfrac{{2{b^2}}}{a} = 36$

$ \Rightarrow {b^2} = 18a$

Now, ${c^2} = {a^2} + {b^2}$

$ \Rightarrow {a^2} = {c^2} - {b^2}$

$ \Rightarrow {a^2} = 144 - 18a$

$ \Rightarrow {a^2} + 18a - 144 = 0$

$ \Rightarrow (a + 24)(a - 6) = 0$

$ \Rightarrow a = 6$ because $a$ is non-negative.

Thus ${a^2} = {6^2} = 36$ & ${b^2} = 108$

Therefore, $\dfrac{{{x^2}}}{{36}} + \dfrac{{{y^2}}}{{108}} = 1$

5. Find the equation of a circle drawn on the diagonal of the rectangle as its diameter, whose sides are $\mathbf{x = 6,{\text{ }}x =  - 3,{\text{ }}y = 3}$ & $\mathbf{y =  - 1}$

Ans: Let ${\text{ABCD}}$ be the given rectangle and $AD = x =  - 3,{\text{ }}BC = x = 6,{\text{ }}AB = y =  - 1$ & $CD = y =  - 3$.

Then $A( - 3, - 1)$ and $C(6,3)$.

The equation of the circle with $AC$ as diameter is:

$(x + 3)(x - 6) + (y + 1)(y - 3) = 0$

$ \Rightarrow {x^2} + {y^2} - 3x - 2y - 21 = 0$

6. Find the coordinates of the focus & vertex, the equations of the directrix & the axis & length of latus rectum of the parabola $\mathbf{{x^2} =  - 8y}$

Ans: ${x^2} =  - 8y$ & ${x^2} =  - 4ay$

So, $4a = 8$

$ \Rightarrow a = 2$

So it is downward parabola.

Foci is $F(0, - a)$ i.e. $F(0, - 2)$.

Vertex is $O(0,0)$.

So, $y = a = 2$.

Its axis is $y - $ axis, whose equation is given by $x = 0$.

Length of latus rectum$ = 4a$ units.

$ = 4 \times 2$ units

$ = 8$ units

7. Show that the equation $\mathbf{6{x^2} + 6{y^2} + 24x - 36y - 18 = 0}$ represents a circle. Also find its centre & radius.

Ans: $6{x^2} + 6{y^2} + 24x - 36y - 18 = 0$

So, ${x^2} + {y^2} + 4x - 6y + 3 = 0$

Where, $2g = 4,{\text{ }}2f =  - 6$ & $c = 3$

$\therefore g = 2,{\text{ }}f =  - 3$ & $c = 3$

Thus, centre of circle is $( - g, - f) = ( - 2,3)$

Radius of circle $ = \sqrt {4 + 9 + 9}  = \sqrt {20} $

$ = 2\sqrt 5 $ units

8. Find the equation of the parabola with focus at $\mathbf{F(5,0)}$ & directrix is $\mathbf{x =  - 5}$

Ans: $F(5,0)$ lies on the right hand side of origin.

Thus, it is a right hand parabola.

Let the required equation be

${y^2} = 4ax$ & $a = 5$

Hence, ${y^2} = 20x$

9. Find the equation of the hyperbola with center at the origin, length of the transverse axis $\mathbf{6}$ & one focus at $\mathbf{(0,4)}$

Ans: Let its equation be $\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1$

It is clear that $c = 4$.

Length of the transverse axis $ = 6$

$ \Rightarrow 2a = 6$

$ \Rightarrow a = 3$

And, ${c^2} = \left( {{a^2} + {b^2}} \right)$

$ \Rightarrow {b^2} = {c^2} - {a^2}$

$ \Rightarrow {b^2} = 16 - 9$

$ \Rightarrow {b^2} =  7$

Thus, ${a^2} = 9$ & ${b^2} =  7$

Hence, equation is $\dfrac{{{x^2}}}{{9}} - \dfrac{{{x^2}}}{{7}} = 1$

10. Find the equation of an ellipse whose vertices are $\mathbf{(0, \pm 13)}$ & the foci are $\mathbf{(0, \pm 5)}$

Ans: Let the equation be $\dfrac{{{x^2}}}{{{b^2}}} + \dfrac{{{y^2}}}{{{a^2}}} = 1$

& $a = 13$

Let foci be $(0, \pm c)$,

$ \Rightarrow c = 5$

$\therefore {b^2} = {a^2} - {c^2}$

$ \Rightarrow {b^2} = 169 - 25$

$ \Rightarrow {b^2} = 144$

$ \Rightarrow {a^2} = 169$

And ${b^2} = 144$

Thus, equation is $\dfrac{{{x^2}}}{{144}} + \dfrac{{{y^2}}}{{169}} = 1$

11. Find the equation of the ellipse whose foci are $\mathbf{(0, \pm 3)}$ & length of whose major axis is $\mathbf{10}$

Ans:Let the required equation be $\dfrac{{{x^2}}}{{{b^2}}} + \dfrac{{{y^2}}}{{{a^2}}} = 1$

Let ${c^2} = {a^2} - {b^2}$

Foci are $(0, \pm c)$ & $c = 3$

$a = $length of the semi- major axis i.e. $\dfrac{1}{2} \times 10 = 5$

So, ${c^2} = {a^2} - {b^2}$

$ \Rightarrow {b^2} = {a^2} - {c^2}$

$ \Rightarrow {b^2} = 25 - 3$

$ \Rightarrow {b^2} = 16$

Thus, ${a^2} = 25$ & ${b^2} = 16$

So, the required equation is $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{{25}} = 1$.

12. Find the equation of the hyperbola with centre at the origin, length of the transverse axis $\mathbf{8}$ & one focus at $\mathbf{(0,6)}$

Ans: Let its equation by $\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1$

It is clear that $c = 6$ 

And the length of the transverse axis $ = 8$

$ \Rightarrow 2a = 8$

$ \Rightarrow a = 4$

And, ${c^2} = {a^2} + {b^2}$

$ \Rightarrow {b^2} = {c^2} - {a^2}$

$ \Rightarrow 36 - 16 = 20$

So, ${a^2} = 16$ & ${b^2} = 20$

So, the required equation is $\dfrac{{{y^2}}}{{16}} - \dfrac{{{x^2}}}{{20}} = 1$

13. Find the equation of the hyperbola whose foci are at $\mathbf{(0, \pm B)}$ & the length of whose conjugate axis is $\mathbf{2\sqrt {11} }$

Ans: Let equation be $\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1$

Let foci be $(0, \pm c)$

$\therefore c = 8$

Length of conjugate axis $ = 2\sqrt {11} $

$ \Rightarrow 2b = 2\sqrt {11} $

$ \Rightarrow b = \sqrt {11} $

$ \Rightarrow {b^2} = 11$

And, ${c^2} = \left( {{a^2} + {b^2}} \right)$

$ \Rightarrow {a^2} = \left( {{c^2} - {b^2}} \right)$

$ \Rightarrow {a^2} = 64 - 11$

$ \Rightarrow {a^2} = 53$

So, the required equation is $\dfrac{{{y^2}}}{{53}} - \dfrac{{{x^2}}}{{11}} = 1$

14. Find the equation of the hyperbola whose vertices are $\mathbf{(0, \pm 3)}$ & foci are $\mathbf{(0, \pm 8)}$

Ans: Vertices are $(0 \pm a)$

It is given that the vertices are $(0 \pm 3)$

$\therefore a = 3$

Let foci be $(0, \pm c)$

It is given that the foci are $(0, \pm 8)$

$\therefore c = 8$

And ${b^2} = \left( {{c^2} - {a^2}} \right)$

$ \Rightarrow {b^2} = {8^2} - {3^2}$

$ \Rightarrow {b^2} = 64 - 9$

$ \Rightarrow {b^2} = 55$

Now, ${a^2} = {3^2} = 9$ & ${b^2} = 55$.

So, the required equation is $\dfrac{{{y^2}}}{9} - \dfrac{{{x^2}}}{{55}} = 1$

15. Find the equation of the ellipse for which $\mathbf{e = \dfrac{4}{5}}$ & whose vertices are $\mathbf{(0. \pm 10)}$.

Ans: Vertices are $(0, \pm a)$ 

So, $a = 10$

Let ${c^2} = \left( {{a^2} - {b^2}} \right)$

$ \Rightarrow e = \dfrac{c}{a}$

$ \Rightarrow c = ae$

$ \Rightarrow c = \left[ {10 \times \dfrac{4}{5}} \right]$

$ \Rightarrow c = 8$

And, ${c^2} = \left( {{a^2} - {b^2}} \right)$

$ \Rightarrow {b^2} = \left( {{a^2} - {c^2}} \right)$

$ \Rightarrow {b^2} = (100 - 64)$

$ \Rightarrow {b^2} = 36$

$\therefore {a^2} = {(10)^2} = 100$ & ${b^2} = 36$

So, the required equation is $\dfrac{{{x^2}}}{{36}} + \dfrac{{{y^2}}}{{100}} = 1$

16. Find the equation of the parabola with vertex at the origin & $\mathbf{{\text{y}} + 5 = 0}$ as its directrix. Also, find its focus

Ans: Let the vertex of the parabola be $O(0,0)$.

$y + 5 = 0$

$ \Rightarrow y =  - 5$

The directrix is a line parallel to the $x - $axis at a distance of $5$ units below the $x - $axis. Thus, the focus is $F(0,5)$.

So, the equation of the parabola is ${x^2} = 4ay$ where $a = 5$ i.e. ${x^2} = 20y$.

17. Find the equation of a circle, the end points of one of whose diameters are $\mathbf{A(2, - 3)}$& $\mathbf{B( - 3,5)}$

Ans: Let the end points of one of whose diameters are $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is given by

$\left( {x - {x_1}} \right)\left( {x - {x_2}} \right) + \left( {y - {y_1}} \right)\left( {y - {y_2}} \right) = 0$

So, ${x_1} = 2,{\text{ }}{y_1} =  - 3$ & ${x_2} =  - 3,{\text{ }}{y_2} = 5$.

$\therefore $ The required equation of the circle is $(x - 2)(x + 3) + (y + 3)(y - 5) = 0$

$ \Rightarrow {x^2} + {y^2} + x - 2y - 21 = 0$

18. Find the equation of ellipse whose vertices are $\mathbf{(0, \pm 13)}$ & the foci are $\mathbf{(0, \pm 5)}$

Ans: Let the required equation be $\dfrac{{{x^2}}}{{{b^2}}} + \dfrac{{{y^2}}}{{{a^2}}} = 1$.

Its vertices are $(0 \pm a)$.

So, $a = 13$

Let its foci be $(0 \pm c)$ then $c = 5$.

$\therefore {b^2} = {a^2} - {c^2}$

$ \Rightarrow {b^2} = 169 - 25$

$ \Rightarrow {b^2} = 144$

Thus ${b^2} = 144$ and ${a^2} = 169$.

So, the required equation is $\dfrac{{{x^2}}}{{144}} + \dfrac{{{y^2}}}{{169}} = 1$.

19. Find the equation of the hyperbola whose foci are $\mathbf{( \pm 5,0)}$ & the transverse axis is of length $\mathbf{8}$.

Ans: Let the required equation be $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$

Length of transverse axis$ = 2a$.

$\therefore 2a = 8$

$ \Rightarrow a = 4$

$ \Rightarrow {a^2} = 16$

Let its foci be $( \pm c,0)$.

So, $c = 5$.

$\therefore {b^2} = \left( {{c^2} - {a^2}} \right)$

$ \Rightarrow {b^2} = {5^2} - {4^2}$

$ \Rightarrow {b^2} = 9$

Thus ${a^2} = 16$ and ${b^2} = 9$

Therefore, the required equation is $\dfrac{{{x^2}}}{{16}} - \dfrac{{{y^2}}}{9} = 1$.

20. Find the equation of a circle, the end points of one of whose diameters are $\mathbf{A( - 3,2)}$& $\mathbf{B(5, - 3)}$

Ans: Let the equation be $\left( {x - {x_1}} \right)\left( {x - {x_2}} \right) + \left( {y - {y_1}} \right)\left( {y - {y_2}} \right) = 0$

So ${x_1} =  - 3,{\text{ }}{y_1} = 2$ and ${x_2} = 5,{\text{ }}{y_2} =  - 3$.

$ \Rightarrow (x + 3)(x - 5) + (y - 2)(y + 3) = 0$

$ \Rightarrow {x^2} - 2x - 15 + {y^2} + y - 6 = 0$

$ \Rightarrow {x^2} + {y^2} - 2x + y - 21 = 0$

21. If eccentricity is $\mathbf{\dfrac{1}{5}}$ & foci are $\mathbf{( \pm 7,0)}$ find the equation of an ellipse.

Ans: Let the required equation of the ellipse be $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$.

Let its foci be $( \pm c,0)$.

So, $c = 7$.

And $e = \dfrac{c}{a}$

$ \Rightarrow a = \dfrac{c}{e}$

$ \Rightarrow a = \dfrac{7}{{\dfrac{1}{5}}}$

$ \Rightarrow a = 35$

Also ${c^2} = \left( {{a^2} - {b^2}} \right)$

$ \Rightarrow {b^2} = {a^2} - {c^2}$

$ \Rightarrow {b^2} = {(35)^2} - 49$

$ \Rightarrow {b^2} = 1225 - 49$

$ \Rightarrow {b^2} = 1176$

$\therefore {a^2} = 1225$ and ${b^2} = 1176$.

So, the required equation is $\dfrac{{{x^2}}}{{1225}} + \dfrac{{{y^2}}}{{1176}} = 1$

22. Find the equation of the hyperbola whose foci are $\mathbf{( \pm 5,0)}$ & the transverse axis is of length

Ans: Let the required equation be $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$

Length of its transverse axis $ = 2a$

$\therefore 2a = 8$

$ \Rightarrow a = 4$

$ \Rightarrow {a^2} = 16$

Let its foci be $( \pm c,0)$

So, $c = 5$.

$\therefore {b^2} = {c^2} - {a^2}$

$ \Rightarrow {b^2} = 25 - 16$

$ \Rightarrow {b^2} = 9$

So, the required equation is $\dfrac{{{x^2}}}{{16}} - \dfrac{{{y^2}}}{9} = 1$

23. Find the length of axes & coordinates of the vertices of the hyperbola $\mathbf{\dfrac{{{x^2}}}{{49}} - \dfrac{{{y^2}}}{{64}} = 1}$

Ans: The equation of the given hyperbola is $\dfrac{{{x^2}}}{{49}} - \dfrac{{{y^2}}}{{64}} = 1$

Compare the given equation with $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$.

$ \Rightarrow {a^2} = 49$ and ${b^2} = 64$.

$\therefore {c^2} = \left( {{a^2} + {b^2}} \right)$

$ \Rightarrow {c^2} = 49 + 64$

$ \Rightarrow {c^2} = 113$

Length of transverse axis is $2a = 2 \times 7 = 14$ units

Length of conjugate axis is $2b = 2 \times 8 = 16$ units

The coordinates of the vertices are $A( - a,0)$ and $B(a, 0)$ i.e. $A( - 7,0)$ and $B(7,0)$.

24. Find the lengths of axes & length of latus rectum of the hyperbola, $\mathbf{\dfrac{{{y^2}}}{9} - \dfrac{{{x^2}}}{{16}} = 1}$

Ans: The given equation is $\dfrac{{{y^2}}}{9} - \dfrac{{{x^2}}}{{16}} = 1$.

Compare the given equation with $\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1$.

$ \Rightarrow {a^2} = 9$ & ${b^2} = 16$

Length of transverse axis is $2a = 2 \times 3 = 6$ units

Length of conjugate axis is $2b = 2 \times 4 = 8$ units

The coordinates of the vertices are $A(0, - a)$ and $B(0,a)$ i.e. $A(0, - 3)$ and $B(0,3)$.

25. Find the eccentricity of the hyperbola of $\mathbf{\dfrac{{{y^2}}}{9} - \dfrac{{{x^2}}}{{16}} = 1}$

Ans: Here, $a = 3$ and $b = 4$

And ${c^2} = {a^2} + {b^2}$

$ \Rightarrow {c^2} = 9 + 16$

$ \Rightarrow {c^2} = 25$

Thus, $c = 5$

$ \Rightarrow e = \dfrac{c}{a}$

$ \Rightarrow e = \dfrac{5}{3}$

26. Find the equation of the ellipse, the ends of whose major axis are $\mathbf{( \pm 3,0)}$ & at the ends of whose minor axis are $\mathbf{(0, \pm 4)}$

Ans: Let the required equation be $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$

Its vertices are $( \pm a,0)$.

So, $a = 3$.

Ends of minor axis are $C(0, - 4)$ and $D(0,4)$.

$\therefore CD = 8$ i.e. length of the minor axis$ = 8$ units

$ \Rightarrow 2b = 8$

$ \Rightarrow b = 4$

$\therefore a = 3$ and $b = 4$.

Therefore, the required equation is $\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{{16}} = 1$.

27. Find the equation of the parabola with focus at $\mathbf{F(4,0)}$ & directrix $\mathbf{x =  - 3}$

Ans: Focus $F(4,0)$ lies on the axis hand side of the origin.

Thus, it is a right handed parabola.

Let the required equation be ${y^2} = 4ax$.

Then, $a = 4$.

So, the required equation is ${y^2} = 16x$.

28. If $\mathbf{y = 2x}$ is a chord of the circle $\mathbf{{x^2} + {y^2} - 10x = 0}$, find the equation of the circle with this chord as a diameter

Ans: $y = 2x$ and ${x^2} + {y^2} - 10x = 0$

Put $y = 2x$ in ${x^2} + {y^2} - 10x = 0$.

$ \Rightarrow 5{x^2} - 10x = 0$

$ \Rightarrow 5x(x - 2) = 0$

$ \Rightarrow x = 0$ or $x = 2$

Now, $x = 0 \Rightarrow y = 0$ and $x = 2 \Rightarrow y = 4$.

$\therefore $ The points of intersection of the given chord and the given circle are$A(0,0)$ and $B(2,4)$.

$\therefore $ The required equation of the circle with ${\text{AB}}$ as diameter is $(x - 0)(x - 2) + (y - 0)(y - 4) = 0$

$ \Rightarrow {x^2} + {y^2} - 2x - 4y = 0$

Very Long Answer Questions: (6 Marks)

1. Find the length of major & minor axis- coordinates of vertices & the foci, the eccentricity & length of latus rectum of the ellipse $\mathbf{16{x^2} + {y^2} = 16}$

Ans: $16{x^2} + {y^2} = 16$

Divide by $16$,

$ \Rightarrow {x^2} + \dfrac{{{y^2}}}{{16}} = 1$

i.e. ${b^2} = 1$ and ${a^2} = 16$

So, $b = 1$ & $a = 4$.

And $c = \sqrt {{a^2} - {b^2}} $

$ \Rightarrow c = \sqrt {16 - 1} $

$ \Rightarrow c = \sqrt {15} $

So, $a = 4,{\text{ }}b = 1$ & $c = \sqrt {15} $.

(i) Length of major axis$ = 2a = 2 \times 4 = 8$ units

Length of minor axis$ = 2b = 2 \times 1 = 2$ units

(ii) Coordinates of the vertices are $A( - a,0)$ & $B(a,0)$ i.e. $A( - 4,0)$ & $B(4,0)$

(iii) Coordinates of foci are ${F_1}( - c,0)$ & ${F_2}(c,0)$ i.e. ${F_1}( - \sqrt {15} ,0)$ & ${F_2}(\sqrt {15},0 )$

(iv) Eccentricity, $e = \dfrac{c}{a} = \dfrac{{\sqrt {15} }}{4}$

(v) Length of latus rectum$ = \dfrac{{2{b^2}}}{a} = \dfrac{2}{4} = \dfrac{1}{2}$ units

2. Find the lengths of the axis, the coordinates of the vertices & the foci the eccentricity & length of the latus rectum of the hyperbola $\mathbf{25{x^2} - 9{y^2} = 225}$

Ans: \[25{x^2} - 9{y^2} = 225\]

\[ \Rightarrow \dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{{25}} = 1\]

Now, ${a^2} = 9$ & ${b^2} = 25$

And $c = \sqrt {{a^2} + {b^2}} $

$ \Rightarrow c = \sqrt {9 + 25} $

$ \Rightarrow c = \sqrt {34} $

(i) Length of transverse axis $ = 2a = 2 \times 3 = 6$ units

Length of conjugate axis $ = 2b = 2 \times 5 = 10$ units

(ii) The coordinates of vertices are $A( - a,0)$ & $B(a,0)$ i.e. $A( - 3,0)$ & $B(3,0)$

(iii) The coordinates of foci are ${F_1}( - c,0)$ & ${F_2}(c,0)$ i.e. ${F_1}( - \sqrt {34} ,0)$ & ${F_2}(\sqrt {34} ,0)$

(iv) Eccentricity, $e = \dfrac{c}{a} = \dfrac{{\sqrt {34} }}{3}$

(v) Length of the latus rectum $ = \dfrac{{2{b^2}}}{a} = \dfrac{{50}}{3}$ units

3. A man running in a race course notes that the sum of the distances of the two flag posts from him is always $\mathbf{12{\text{ m}}}$& the distance between the flag posts is $\mathbf{10{\text{ m}}}$. Find the equation of the path traced by the man.

Ans: Ellipse is the locus of a point that moves in such a way that the sum of its distances from two fixed points is constant.

Thus, the path is an ellipse.

Let the equation of the ellipse be $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$.

Where ${b^2} = {a^2}\left( {1 - {c^2}} \right)$

It is clear that $2a = 12$ & $2ae = 10$

$ \Rightarrow a = b$ and $e = \dfrac{5}{6}$

$ \Rightarrow {b^2} = {a^2}\left( {1 - {e^2}} \right)$

$ \Rightarrow {b^2} = 36\left( {1 - \dfrac{{25}}{{36}}} \right)$

$ \Rightarrow {b^2} = 11$

So, the required equation is $\dfrac{{{x^2}}}{{36}} + \dfrac{{{y^2}}}{{11}} = 1$.

4. An equilateral triangle is inscribed in the parabola $\mathbf{{y^2} = 4ax}$ so that one angular point of the triangle is at the vertex of the parabola. Find the length of each side of the triangle.

Ans: Let $OPQ$ be an equilateral triangle inscribed in the parabola ${y^2} = 4ax$ where $O(0,0)$ is the vertex so that $\angle POM = \angle QOM = {30^\circ }$.

Let $OP = OQ = r$.

And $P = (r\cos {30^ \circ },r\sin {30^ \circ })$

$ \Rightarrow P = \left( {\dfrac{{r\sqrt 3 }}{2},\dfrac{r}{2}} \right)$

$P$ lies on parabola. 

$ \Rightarrow \dfrac{{{r^2}}}{4} = 4a\left( {\dfrac{{r\sqrt 3 }}{2}} \right)$

$ \Rightarrow r = 8a\sqrt 3 $

Hence, the length of each side of the triangle is $8a\sqrt 3 $ units.

5. Find the equation of the hyperbola whose foci are at $\mathbf{(0, \pm \sqrt {10} )}$ & which passes through the points $\mathbf{(2,3)}$

Ans: Let equation be \[\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1{\text{      }}...{\text{(}}i)\]

Let foci be $(0, \pm c)$.

But the foci are $(0, \pm \sqrt {10} )$.

$\therefore c = \sqrt {10} $

$ \Rightarrow {c^2} = 10$

$ \Rightarrow \left( {{a^2} + {b^2}} \right) = 10{\text{    }}...{\text{(}}ii)$

As \[(i)\] passes through $(2,3)$,

$ \Rightarrow \dfrac{9}{{{a^2}}} - \dfrac{4}{{{b^2}}} = 1$

$ \Rightarrow \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 {a^4} - 23{a^2} + 90 = 0$

$ \Rightarrow \left( {{a^2} - 18} \right)\left( {{a^2} - 5} \right) = 0$

$ \Rightarrow {a^2} = 5$

$\because {a^2} = 18 \Rightarrow {b^2} =  - 8$, which is impossible

So, ${a^2} = 5$ and ${b^2} = 5$.

So, the required equation is $\dfrac{{{y^2}}}{5} - \dfrac{{{x^2}}}{5} = 1$,

i.e. ${y^2} - {x^2} = 5$.

6. Find the equation of the curve formed by the set of all these points the sum of whose distance from the points $\mathbf{A(4,0,0)}$ & $\mathbf{B( - 4,0,0)}$ is $\mathbf{10}$ units.

Ans: Let $P(x,y,z)$ be an arbitrary point on the given curve.

So, $PA + PB = 10$

$ \Rightarrow \sqrt {{{(x - 4)}^2} + {y^2} + {z^2}}  + \sqrt {{{(x + 4)}^2} + {y^2} + {z^2}}  = 10$

$ = \sqrt {{{(x + 4)}^2} + {y^2} + {z^2}}  = 10 - \sqrt {{{(x - 4)}^2} + {y^2} + {z^2}} $

Squaring both sides:

$ \Rightarrow {(x + 4)^2} + {y^2} + {z^2} = 100 + {(x - 4)^2} + {y^2} + {z^2} - 20\sqrt {{{(x - 4)}^2} + {y^2} + {z^2}} $

$ \Rightarrow 16x = 100 - 20\sqrt {{{(x - 4)}^2} + {y^2} + {z^2}} $

$ \Rightarrow 5\sqrt {{{(x - 4)}^2} + {y^2} + {z^2}}  = 25 - 4x$

$ \Rightarrow 25\left[ {{{(x - 4)}^2} + {y^2} + {z^2}} \right] = 625 + 16{x^2} - 200x$

$ \Rightarrow 9{x^2} + 25{y^2} + 25{z^2} - 225 = 0$

So, the required equation of the curve is $9{x^2} + 25{x^2} + 25{z^2} - 225 = 0$.

7. Find the equation of the ellipse with centre at the origin, major axis on the $\mathbf{y - } $axis & passing through the points $\mathbf{(3,2)}$ & $\mathbf{(1,6)}$

Ans: Let the required equation be $\dfrac{{{x^2}}}{{{b^2}}} + \dfrac{{{y^2}}}{{{a^2}}} = 1{\text{         }}...(i)$

Since $(3,2)$ lies on $(i)$, $\dfrac{9}{{{b^2}}} + \dfrac{4}{{{a^2}}} = 1{\text{             }} \ldots (ii)$

As $(1,6)$ lies on $(i)$, $\dfrac{1}{{{b^2}}} + \dfrac{{36}}{{{a^2}}} = 1{\text{                }} \ldots (iii)$

Put $\dfrac{1}{{{b^2}}} = u$ & $\dfrac{1}{{{a^2}}} = v$.

$ \Rightarrow 9u + 4v = 1{\text{            }} \ldots (iv)$ and $u + 36v = 1{\text{             }} \ldots (v)$

Multiply $(v)$ by $9$ & subtract $(iv)$ from it.

$ \Rightarrow 320v = 8$

$ \Rightarrow v = \dfrac{8}{{320}}$

$ \Rightarrow v = \dfrac{1}{{40}}$

$ \Rightarrow \dfrac{1}{{{a^2}}} = \dfrac{1}{{40}}$

$ \Rightarrow {a^2} = 40$

Put $v = \dfrac{1}{{40}}$ in $(v)$

$ \Rightarrow u + \left[ {36 \times \dfrac{1}{{40}}} \right] = 1$

$ \Rightarrow u = \left[ {1 - \dfrac{9}{{10}}} \right] = \dfrac{1}{{10}}$

$ \Rightarrow \dfrac{1}{{{b^2}}} = \dfrac{1}{{10}}$

$ \Rightarrow {b^2} = 10$

So, ${b^2} = 10$ and ${a^2} = 40$

So, the required equation is $\dfrac{{{x^2}}}{{10}} + \dfrac{{{y^2}}}{{40}} = 1$.

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