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

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.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 10 Exercise 10.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

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