NCERT Solutions Class 11 Maths Chapter 10 Conic Sections Exercise 10.3

Exercise 10.3

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 ${{x}^{2}}+4{{y}^{2}}=52$. 

Class 11 Maths Chapter 10: Exercises Breakdown

Conclusion

NCERT Solutions for Class 11 Maths Chapter 10 - Conic Sections: Exercise 10.3 provide a thorough understanding of ellipses, covering important concepts such as the standard equation, eccentricity, and the positions of foci. These step-by-step solutions simplify complex problems and make it easier for students to grasp the key ideas and solve exam-related questions with confidence. By using these solutions, students can enhance their understanding of ellipses and strengthen their problem-solving skills. Don't forget to download the FREE PDF for easy access and effective exam preparation!

CBSE Class 11 Maths Chapter 10 - Conic Sections Other Study Materials

Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Additional Study Materials for Class 11 Maths