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# NCERT Solutions for Class 11 Maths Chapter 10: Conic Sections - Exercise 10.3

Last updated date: 11th Sep 2024
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## NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 (Ex 10.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 10 Conic Sections Exercise 10.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

 Class: NCERT Solutions for Class 11 Subject: Class 11 Maths Chapter Name: Chapter 10 - Conic Sections Exercise: Exercise - 10.3 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes
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## Access NCERT Solutions for Class 11 Maths Chapter 10- Conic Sections

Exercise 10.3

Refer the page 12 to 22 for exercise 10.3 in the PDF

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 obtain $a = 6$ and $b = 4$.

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

As we know the value of $a$ and $b$

$= \sqrt {36 - 16}$

$= \sqrt {20}$

$= \sqrt {4 \times 5}$

$= 2\sqrt 5$

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, $2a = 12$

Dividing both side by 2 we get $a = 6$

Length of minor axis, $2b = 8$

Dividing both side by 2 we get $b = 4$

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 obtain $b = 2$ and $a = 5$

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

As we know the value of $a$ and $b$

$= \sqrt {25 - 4}$

$= \sqrt {21}$

Therefore,

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

The coordinates of the vertices are $\left( {0,5} \right)$ and $\left( {0,-5} \right)$

Length of major axis,$2a = 10$

Dividing both side by 2 we get $a = 5$

Length of minor axis, $2b = 4$

Dividing both side by 2 we get $b = 2$

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 minor axis, the major axis, the eccentricity and the length of the latus rectum of the ellipse $\dfrac{{{x^2}}}{{{4^2}}} + \dfrac{{{y^2}}}{{{3^2}}} = 1$

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 minor axis is along the $y -$axis, while the major axis is along the $x -$axis

On comparing the given equation with $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ , we obtain $a = 4$ and $b = 3$

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

As we know the value of $a$ and $b$,

$= \sqrt {16 - 9}$

$= \sqrt 7$

Therefore,

The coordinates of the foci are $\left( {\sqrt 7 ,0} \right)$ and $\left( { - \sqrt 7 ,0} \right)$ .

The coordinates of the vertices are $\left( {4,0} \right)$ and $\left( { - 4,0} \right)$

Length of major axis,$2a = 8$

Dividing both side by 2 we get, $a = 4$

Length of minor axis, $2b = 6$

Dividing both side by 2 we get $b = 3$

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 minor axis, the major axis, the eccentricity and the length of the latus rectum of the ellipse $\dfrac{{{x^2}}}{{{{25}^2}}} + \dfrac{{{y^2}}}{{{{100}^2}}} = 1$

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 obtain $a = 10$ and $b = 5$

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

As we know the value of $a$ and $b$

$= \sqrt {100 - 25}$

$= \sqrt {75}$

$= \sqrt {25 \times 3}$

$= 5\sqrt 3$

Therefore,

The coordinates of the foci are $\left( {0,5\sqrt 3 } \right)$ and $\left( {0, - 5\sqrt 3 } \right)$ .

The coordinates of the vertices are $\left( {0,10} \right)$ and$(0, - 10)$.

Length of major axis,$2a = 20$

Dividing both side by 2 we get, $a = 10$

Length of minor axis, $2b = 10$

Dividing both side by 2 we get, $b = 5$

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 obtain $a = 7$ and $b = 6$

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

As we know the value of $a$ and $b$,

$= \sqrt {49 - 36}$

$= \sqrt {13}$

Therefore,

The coordinates of the foci are $\left( {\sqrt {13} ,0} \right)$ and $\left( { - \sqrt {13} ,0} \right)$ .

The coordinates of the vertices are $\left( {7,0} \right)$ and $\left( { - 7,0} \right)$.

Length of major axis,$2a = 14$

Dividing both side by 2 we get $a = 7$

Length of minor axis, $2b = 12$

Dividing both side by 2 we get $b = 6$

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 $\dfrac{{{y^2}}}{{{{20}^2}}}$ is greater than the denominator of$\dfrac{{{x^2}}}{{{{10}^2}}}$.

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 obtain $a = 20$ and $b = 10$

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

As we know the value of $a$ and $b$,

$= \sqrt {400 - 100}$

$= \sqrt {300}$

$= 10\sqrt 3$

Therefore,

The coordinates of the foci are $\left( {0,10\sqrt 3 } \right)$ and $\left( {0, - 10\sqrt 3 } \right)$ .

The coordinates of the vertices are $\left( {0,20} \right)$ and $\left( {0, - 20} \right)$

Length of major axis,$2a = 40$

Dividing both side by 2 we get $a = 20$

Length of minor axis, $2b = 20$

Dividing both side by 2 we get $b = 10$

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 $36{x^2}{\text{ }} + {\text{ }}4{y^2}{\text{ }} = {\text{ }}144$.

It can be written as

$36{x^2}{\text{ }} + {\text{ }}4{y^2}{\text{ }} = {\text{ }}144$

Dividing both side by 144 we get

$\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{{36}} = 1$

$\dfrac{{{x^2}}}{{{2^2}}} + \dfrac{{{y^2}}}{{{6^2}}} = 1$

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 obtain $b = 2$ and$a = 6$.

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

As we know the value of $a$ and $b$,

$= \sqrt {36 - 4}$

$= \sqrt {32}$

$= 4\sqrt 2$

Therefore, the coordinates of the foci are $\left( {0,4\sqrt 2 } \right)$ and $\left( {0, - 4\sqrt 2 } \right)$.

The coordinates of the vertices are $\left( {0, \pm 6} \right)$

Length of major axis,$2a = 12$

Dividing both side by 2 we get $a = 6$

Length of minor axis, $2b = 4$

Dividing both side by 2 we get $b = 2$

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 $16{x^2}{\text{ }} + {\text{ }}{y^2}{\text{ }} = {\text{ }}16$.

It can be written as

$16{x^2}{\text{ }} + {\text{ }}{y^2}{\text{ }} = {\text{ }}16$

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

$\dfrac{{{x^2}}}{{{1^2}}} + \dfrac{{{y^2}}}{{{4^2}}} = 1$

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 obtain $b = 1$ and$a = 4$.

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

$= \sqrt {16 - 1}$

$= \sqrt {15}$

Therefore, the coordinates of the foci are $\left( {0,\sqrt {15} } \right)$ and $\left( {0, - \sqrt {15} } \right)$.

The coordinates of the vertices are $\left( {0, \pm 4} \right)$

Length of major axis, $2a = 8$

Dividing both side by 2 we get $a = 4$

Length of minor axis, $2b = 2$

Dividing both side by 2 we get, $b = 1$

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

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

$4{x^2} + 9{y^2} = 36$

$\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{4} = 1$

$\dfrac{{{x^2}}}{{{3^2}}} + \dfrac{{{y^2}}}{{{2^2}}} = 1$

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 obtain $b = 2$ and$a = 3$.

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

$= \sqrt {9 - 4}$

$= \sqrt 5$

Therefore, the coordinates of the foci are $\left( {\sqrt 5 ,0} \right)$ and $\left( { - \sqrt 5 ,0} \right)$.

The coordinates of the vertices are $\left( { \pm 3,0} \right)$

Length of major axis, $2a = 6$

Dividing both side by 2 we get $a = 3$

Length of minor axis, $2b = 4$

Dividing both side by 2 we get $b = 2$

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 $\left( { \pm {\mathbf{5}},{\text{ }}{\mathbf{0}}} \right)$, foci$( \pm {\mathbf{4}},{\text{ }}{\mathbf{0}})$.

Ans: Vertices $\left( { \pm {\mathbf{5}},{\text{ }}{\mathbf{0}}} \right)$, foci  $( \pm {\mathbf{4}},{\text{ }}{\mathbf{0}})$

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

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

As we know the value of $\;a$ and $c$

$\therefore {5^2} = {b^2} + {4^2}$

$25 = {b^2} + 16$

${b^2} = 25 - 16$

$b = \sqrt 9$

$b = 3$

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 $\left( {0, \pm 13} \right)$, foci$(0, \pm 5)$.

Ans: Vertices $\left( {0, \pm 13} \right)$, 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

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

As we know the value of $\;a$ and $c$

$\therefore {13^2} = {b^2} + {5^2}$

$169 = {b^2} + 25$

${b^2} = 169 - 25$

$b = \sqrt {144}$

$b = 12$

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 $( \pm 6,{\text{ 0}})$, foci$( \pm {\mathbf{4}},{\text{ }}{\mathbf{0}})$.

Ans: Vertices $( \pm 6,{\text{ 0}})$, foci  $( \pm {\mathbf{4}},{\text{ }}{\mathbf{0}})$

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

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

As we know the value of $\;a$ and $c$

$\therefore {6^2} = {b^2} + {4^2}$

$36 = {b^2} + 16$

${b^2} = 36 - 16$

$b = \sqrt {20}$

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 $( \pm 3,{\text{ 0}})$, ends of minor axis$(0, \pm {\mathbf{2}})$.

Ans: Here we have, ends of major axis $( \pm 3,{\text{ 0}})$ and ends of minor axis$(0, \pm {\mathbf{2}})$.

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 semi-major 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: Here we have, ends of major axis $(0, \pm \sqrt 5 )$ and 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 semi-major 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 $( \pm 5,{\text{ 0}})$

Ans: Length of major axis = 26; foci = $( \pm 5,{\text{ 0}})$.

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 semi-major axis.

Accordingly,

$2a = 26 \Rightarrow a = 13$ and $c = 5$.

It is known that

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

As we know the value of $\;a$ and $c$

$\therefore {13^2} = {b^2} + {5^2}$

$169 = {b^2} + 25$

${b^2} = 169 - 25$

$b = \sqrt {144}$

$b = 12$

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 semi-major axis.

Accordingly,

$2b = 16 \Rightarrow b = 8$  and  $c = 6$.

It is known that

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

As we know the value of $\;b$ and $c$

$\therefore {a^2} = {8^2} + {6^2}$

${a^2} = 64 + 36$

${a^2} = 100$

$a = \sqrt {100}$

$a = 10$

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 $\left( { \pm {\mathbf{3}},{\text{ }}{\mathbf{0}}} \right)$, $a = 4$

Ans: Foci = $\left( { \pm {\mathbf{3}},{\text{ }}{\mathbf{0}}} \right)$ and $a = 4$.

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 semi-major axis.

Accordingly,

$c = 3$ and $a = 4$.

It is known that

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

As we know the value of $\;a$ and $c$

$\therefore {4^2} = {b^2} + {3^2}$

$16 = {b^2} + 9$

${b^2} = 16 - 9$

$b = \sqrt 7$

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:${\mathbf{b}} = {\mathbf{3}}, {\mathbf{c}} = {\mathbf{4}}$, centre at the origin; foci on the ${\mathbf{x}} -$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 semi-major axis.

Accordingly,

$b = 3$ and $c = 4$.

It is known that

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

As we know the value of $\;b$ and $c$

$\therefore {a^2} = {3^2} + {4^2}$

${a^2} = 9 + 16$

${a^2} = 25$

$a = \sqrt {25}$

$a = 5$

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 $\left( {0,0} \right)$, major axis on the y-axis and passes through the points $\left( {{\mathbf{3}},{\mathbf{2}}} \right)$ and $\left( {{\mathbf{1}},{\mathbf{6}}} \right)$

Ans: Since the centre is at $\left( {0,0} \right)$ 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 $\left( {{\mathbf{3}},{\mathbf{2}}} \right)$ and $\left( {{\mathbf{1}},{\mathbf{6}}} \right)$. 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 obtain ${b^2} = {\text{ }}10$ and${a^2} = {\text{ }}40$.

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 $\left( {4,3} \right)$ and $\left( {6,2} \right)$.

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

$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$

Where,$a$ is the semi-major axis.

The ellipse passes through points $\left( {4,3} \right)$ and $\left( {6,2} \right)$. Hence,

$\dfrac{{16}}{{{a^2}}} + \dfrac{9}{{{b^2}}} = 1$

$\dfrac{{36}}{{{a^2}}} + \dfrac{4}{{{b^2}}} = 1$

On solving above equations, we obtain ${a^2} = {\text{ 52}}$ and ${b^2} = {\text{ 13}}$.

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

## NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections Exercise 10.3

Opting for the NCERT solutions for Ex 10.3 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 10.3 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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## FAQs on NCERT Solutions for Class 11 Maths Chapter 10: Conic Sections - Exercise 10.3

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2. How many questions are there in NCERT Solutions for Class 11 Maths Chapter 10 Conic sections (Ex 10.3) Exercise 10.3?

In Exercise 10.3 of chapter 10 Conic Sections contains a total of 20 questions. NCERT questions can be considered a useful resource as it contains lots of strategies that will help students to achieve better results. Vedantu provides NCERT Solutions for all the chapters of Class 11 maths free of cost to help the students in their exam preparation. On the official website of Vedantu and the mobile application of Vedantu, you can download these solutions in PDF format.

3. What concepts are being discussed in chapter 10 Conic sections of Class 11 Maths?

NCERT Class 11 Maths discussed the various conic sections and explained them with the help of well-illustrated diagrams. This chapter especially focuses on the descriptions of certain mathematical figures like circles, hyperbolas, parabolas, and ellipses. Students should understand each concept from this chapter very well because there are a lot of questions that are asked in their board final exams and also other competitive exams such as JEE Advance, JEE Mains, etc.

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5. How can I revise Chapter 10 Conic sections of NCERT Class 11 Maths?

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