Exercise 10.3 (Revised) - Chapter 11 - Conic Sections - Ncert Solutions class 11 - Maths
Updated On 26-08-2025 By Lithanya
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Chapter 10: Conic Sections - NCERT Solutions for Class 11 Maths
In each of the Exercises 1 to 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.
1. $\frac{x^2}{36}+\frac{y^2}{16}=1$
Ans. Given: Equation of ellipse: $\frac{x^2}{36}+\frac{y^2}{16}=1$
$\because 36>16$ (it's a horizontal ellipse)
$$
\begin{aligned}
& \therefore a^2=36, b^2=16 \\
& \Rightarrow a=6, b=4
\end{aligned}
$$
Now $c=\sqrt{a^2-b^2}=\sqrt{36-16}=\sqrt{20}=2 \sqrt{5}$
$\therefore$ Coordinates of foci are $( \pm c, 0)$
$$
\Rightarrow( \pm 2 \sqrt{5}, 0)
$$
Coordinates of vertices are $( \pm a, 0)$
$$
\Rightarrow( \pm 6,0)
$$
Length of major axis $=2 a=2 \times 6=12$ units
Length of minor axis $=2 b=2 \times 4=8$ units
Eccentricity $(e)=\frac{c}{a}=\frac{2 \sqrt{5}}{6}=\frac{\sqrt{5}}{3}$
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 16}{6}=\frac{16}{3}$ units
2. $\frac{x^2}{4}+\frac{y^2}{25}=1$
Ans. Given: Equation of ellipse: $\frac{x^2}{4}+\frac{y^2}{25}=1$
$\because 25>4$ (it's a vertical ellipse)
$$
\begin{aligned}
& \therefore a^2=25, b^2=4 \\
& \Rightarrow a=5, b=2
\end{aligned}
$$
Now $c=\sqrt{a^2-b^2}=\sqrt{25-4}=\sqrt{21}$
$\therefore$ Coordinates of foci are $(0, \pm c)$
$$
\Rightarrow(0, \pm \sqrt{21})
$$
Coordinates of vertices are $(0, \pm a)$
$$
\Rightarrow(0, \pm 5)
$$
Length of major axis $=2 a=2 \times 5=10$ units
Length of minor axis $=2 b=2 \times 2=4$ units
Eccentricity $(e)=\frac{c}{a}=\frac{\sqrt{21}}{5}$
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 4}{5}=\frac{8}{5}$ units
3. $\frac{x^2}{16}+\frac{y^2}{9}=1$
Ans. Given: Equation of ellipse: $\frac{x^2}{16}+\frac{y^2}{9}=1$
$\because 16>9$ (it's a horizontal ellipse)
$$
\begin{aligned}
& \therefore a^2=16, b^2=9 \\
& \Rightarrow a=4, b=3
\end{aligned}
$$
Now $c=\sqrt{a^2-b^2}=\sqrt{16-9}=\sqrt{7}$
$\therefore$ Coordinates of foci are $( \pm c, 0)$
$$
\Rightarrow( \pm \sqrt{7}, 0)
$$
Coordinates of vertices are $( \pm a, 0)$
$$
\Rightarrow( \pm 4,0)
$$
Length of major axis $=2 a=2 \times 4=8$ units
Length of minor axis $=2 b=2 \times 3=6$ units
Eccentricity $(e)=\frac{c}{a}=\frac{\sqrt{7}}{4}$
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 9}{4}=\frac{9}{2}$ units
4. $\frac{x^2}{25}+\frac{y^2}{100}=1$
Ans. Given: Equation of ellipse: $\frac{x^2}{25}+\frac{y^2}{100}=1$
$\because 100>25$ (it's a vertical ellipse)
$$
\begin{aligned}
& \therefore a^2=100, b^2=25 \\
& \Rightarrow a=10, b=5
\end{aligned}
$$
Now $c=\sqrt{a^2-b^2}=\sqrt{100-25}=\sqrt{75}=5 \sqrt{3}$
$\therefore$ Coordinates of foci are $(0, \pm c)$
$$
\Rightarrow(0, \pm 5 \sqrt{3})
$$
Coordinates of vertices are $(0, \pm a)$
$$
\Rightarrow(0, \pm 10)
$$
Length of major axis $=2 a=2 \times 10=20$ units
Length of minor axis $=2 b=2 \times 5=10$ units
Eccentricity $(e)=\frac{c}{a}=\frac{5 \sqrt{3}}{10}=\frac{\sqrt{3}}{2}$
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 25}{10}=5$ units
5. $\frac{x^2}{49}+\frac{y^2}{36}=1$
Ans. Given: Equation of ellipse: $\frac{x^2}{49}+\frac{y^2}{36}=1$
$\because 49>36$ (it's a horizontal ellipse)
$$
\begin{aligned}
& \therefore a^2=49, b^2=36 \\
& \Rightarrow a=7, b=6
\end{aligned}
$$
Now $c=\sqrt{a^2-b^2}=\sqrt{49-36}=\sqrt{13}$
$\therefore$ Coordinates of foci are $( \pm c, 0)$
$$
\Rightarrow( \pm \sqrt{13}, 0)
$$
Coordinates of vertices are $( \pm a, 0)$
$$
\Rightarrow( \pm 7,0)
$$
Length of major axis $=2 a=2 \times 7=14$ units
Length of minor axis $=2 b=2 \times 6=12$ units
Eccentricity $(e)=\frac{c}{a}=\frac{\sqrt{13}}{7}$
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 36}{7}=\frac{72}{7}$ units
6. $\frac{x^2}{100}+\frac{y^2}{400}=1$
Ans. Given: Equation of ellipse: $\frac{x^2}{100}+\frac{y^2}{400}=1$
$\because 400>100$ (it's a vertical ellipse)
$$
\begin{aligned}
& \therefore a^2=400, b^2=100 \\
& \Rightarrow a=20, b=10
\end{aligned}
$$
Now $c=\sqrt{a^2-b^2}=\sqrt{400-100}=\sqrt{300}=10 \sqrt{3}$
$\therefore$ Coordinates of foci are $(0, \pm c)$
$$
\Rightarrow(0, \pm 10 \sqrt{3})
$$
Coordinates of vertices are $(0, \pm a)$
$$
\Rightarrow(0, \pm 20)
$$
Length of major axis $=2 a=2 \times 20=40$ units
Length of minor axis $=2 b=2 \times 10=20$ units
Eccentricity $(e)=\frac{c}{a}=\frac{10 \sqrt{3}}{20}=\frac{\sqrt{3}}{2}$
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 100}{20}=10$ units
7. $36 x^2+4 y^2=144$
Ans. Given: Equation of ellipse: $36 x^2+4 y^2=144$ $\Rightarrow \frac{36 x^2}{144}+\frac{4 y^2}{144}=1$ (divide both sides of equation by 144 )
$$
\begin{aligned}
& \Rightarrow \frac{x^2}{4}+\frac{y^2}{36}=1 \\
& \because 36>4 \text { (it's a vertical ellipse) } \\
& \therefore a^2=36, b^2=4 \\
& \Rightarrow a=6, b=2
\end{aligned}
$$
Now $c=\sqrt{a^2-b^2}=\sqrt{36-4}=\sqrt{32}=4 \sqrt{2}$
$\therefore$ Coordinates of foci are $(0, \pm c)$
$$
\Rightarrow(0, \pm 4 \sqrt{2})
$$
Coordinates of vertices are $(0, \pm a)$
$$
\Rightarrow(0, \pm 6)
$$
Length of major axis $=2 a=2 \times 6=12$ units
Length of minor axis $=2 b=2 \times 2=4$ units
Eccentricity $(e)=\frac{c}{a}=\frac{4 \sqrt{2}}{6}=\frac{2 \sqrt{2}}{3}$
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 4}{6}=\frac{4}{3}$ units
8. $16 x^2+y^2=16$
Ans. Given: Equation of ellipse: $16 x^2+y^2=16$
$\Rightarrow \frac{16 x^2}{16}+\frac{y^2}{16}=1$ (divide both sides of equation by 16 )
$$
\Rightarrow \frac{x^2}{1}+\frac{y^2}{16}=1
$$
$\because 16$ > 1 (it's a vertical ellipse)
$$
\begin{aligned}
& \therefore a^2=16, b^2=1 \\
& \Rightarrow a=4, b=1
\end{aligned}
$$
Now $c=\sqrt{a^2-b^2}=\sqrt{16-1}=\sqrt{15}$
$\therefore$ Coordinates of foci are $(0, \pm c)$
$$
\Rightarrow(0, \pm \sqrt{15})
$$
Coordinates of vertices are $(0, \pm a)$
$$
\Rightarrow(0, \pm 4)
$$
Length of major axis $=2 a=2 \times 4=8$ units
Length of minor axis $=2 b=2 \times 1=2$ units
Eccentricity $(e)=\frac{c}{a}=\frac{\sqrt{15}}{4}$
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 1}{4}=\frac{1}{2}$ units
9. $4 x^2+9 y^2=36$
Ans. Given: Equation of ellipse: $4 x^2+9 y^2=36$
$\Rightarrow \frac{4 x^2}{36}+\frac{9 y^2}{36}=1$ (divide both sides of equation by 36 )
$\Rightarrow \frac{4 x^2}{36}+\frac{9 y^2}{36}=1$ (divide both sides of equation by 36 )
$$
\Rightarrow \frac{x^2}{9}+\frac{y^2}{4}=1
$$
$\because 9>4$ (it's a horizontal ellipse)
$$
\begin{aligned}
& \therefore a^2=9, b^2=4 \\
& \Rightarrow a=3, b=2
\end{aligned}
$$
Now $c=\sqrt{a^2-b^2}=\sqrt{9-4}=\sqrt{5}$
$\therefore$ Coordinates of foci are $( \pm c, 0)$
$$
\Rightarrow( \pm \sqrt{5}, 0)
$$
Coordinates of vertices are $( \pm a, 0)$
$$
\Rightarrow( \pm 3,0)
$$
Length of major axis $=2 a=2 \times 3=6$ units
Length of minor axis $=2 b=2 \times 2=4$ units
Eccentricity $(e)=\frac{c}{a}=\frac{\sqrt{5}}{3}$
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 4}{3}=\frac{8}{3}$ units
In each of the Exercises 10 to 20, find the equation of the ellipse that satisfies the given conditions:
10. Vertices $( \pm 5,0)$, foci $( \pm 4,0)$
Ans. Since foci $( \pm 4,0)$ lie on $x$-axis, therefore equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Now Vertices $( \pm a, 0)=( \pm 5,0)$
$$
\Rightarrow a=5
$$
And Foci $( \pm c, 0)=( \pm 4,0)$
$$
\Rightarrow c=4
$$
$$
\begin{aligned}
& \because c^2=a^2-b^2 \therefore(4)^2=(5)^2-b^2 \\
& \Rightarrow b^2=25-16=9
\end{aligned}
$$
Therefore, the required equation of ellipse is $\frac{x^2}{25}+\frac{y^2}{9}=1$.
11. Vertices $(0, \pm 13)$, foci $(0, \pm 5)$
Ans. Since foci $(0, \pm 5)$ lie on $y$-axis, therefore equation of ellipse is $\frac{y^2}{a^2}+\frac{x^2}{b^2}=1$
Now Vertices $(0, \pm a)=(0, \pm 13)$
$\Rightarrow a=13$
And Foci $(0, \pm c)=(0, \pm 5)$
$\Rightarrow c=5$
$\because c^2=a^2-b^2$
$\therefore(5)^2=(13)^2-b^2$
$\Rightarrow b^2=169-25=144$
Therefore, the required equation of ellipse is $\frac{x^2}{144}+\frac{y^2}{169}=1$.
12. Vertices $( \pm 6,0)$, foci $( \pm 4,0)$
Ans. Since foci $( \pm 4,0)$ lie on $x$-axis, therefore equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Now Vertices $( \pm a, 0)=( \pm 6,0)$
$$
\Rightarrow a=6
$$
And Foci $( \pm c, 0)=( \pm 4,0)$
$$
\begin{aligned}
& \Rightarrow c=4 \\
& \because c^2=a^2-b^2 \\
& \therefore(4)^2=(6)^2-b^2 \\
& \Rightarrow b^2=36-16=20
\end{aligned}
$$
Therefore, the required equation of ellipse is $\frac{x^2}{36}+\frac{y^2}{20}=1$.
13. Ends of major axis $( \pm 3,0)$, ends of minor axis $(0, \pm 2)$
Ans. Ends of major axis $( \pm 3,0)$ lie on $x$-axis, therefore equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Now Ends of major axis $( \pm a, 0)=( \pm 3,0)$
$$
\Rightarrow a=3
$$
And Ends of minor of axis $(0, \pm b)=(0, \pm 2)$
$$
\Rightarrow b=2
$$
Therefore, the required equation of ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1$.
14. Ends of major axis $(0, \pm \sqrt{5})$, ends of minor axis $( \pm 1,0)$
Ans. Ends of major axis $(0, \pm \sqrt{5})$ lie on $y$-axis, therefore equation of ellipse is $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
Now Ends of major axis $(0, \pm a)=(0, \pm \sqrt{5})$
$$
\Rightarrow a=\sqrt{5}
$$
And Ends of minor of axis $( \pm b, 0)=( \pm 1,0)$
$$
\Rightarrow b=1
$$
Therefore, the required equation of ellipse is $\frac{x^2}{1}+\frac{y^2}{5}=1$.
15. Length of major axis 26 , foci $( \pm 5,0)$
Ans. Since foci $( \pm 5,0)$ lie on $x$-axis, therefore equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\therefore$ Length of major axis $=2 a=26$
$$
\Rightarrow a=13
$$
Foci $( \pm c, 0)=( \pm 5,0)$
$$
\begin{aligned}
& \Rightarrow c=5 \\
& c^2=a^2-b^2 \\
& \Rightarrow(5)^2=(13)^2-b^2
\end{aligned}
$$
$$
\Rightarrow b^2=169-25=144
$$
Therefore, the required equation of ellipse is $\frac{x^2}{169}+\frac{y^2}{144}=1$.
16. Length of minor axis 16 , foci $(0, \pm 6)$
Ans. Since foci $(0, \pm 6)$ lie on $y$-axis, therefore equation of ellipse is $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
$\therefore$ Length of major axis $=2 b=16$
$$
\begin{aligned}
& \Rightarrow b=8 \\
& \text { Foci }(0, \pm c)=(0, \pm 6) \\
& \Rightarrow c=6 \\
& c^2=a^2-b^2 \\
& \Rightarrow(6)^2=a^2-(8)^2 \\
& \Rightarrow a^2=36+64=100
\end{aligned}
$$
Therefore, the required equation of ellipse is $\frac{x^2}{64}+\frac{y^2}{100}=1$.
17. Foci $( \pm 3,0), a=4$
Ans. Since foci $( \pm 3,0)$ lie on $x$-axis, therefore equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$$
\begin{aligned}
& \therefore \text { Foci }( \pm c, 0)=( \pm 3,0) \\
& \Rightarrow c=3 \\
& c^2=a^2-b^2 \\
& \Rightarrow(3)^2=(4)^2-b^2 \\
& \Rightarrow b^2=16-9=7
\end{aligned}
$$
Therefore, the required equation of ellipse is $\frac{x^2}{16}+\frac{y^2}{7}=1$.
18. $b=3, c=4$, centre at origin; foci on $x$-axis
Ans. Since foci lie on $x$-axis, therefore equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$$
\therefore c^2=a^2-b^2
$$
$$
\begin{aligned}
& \Rightarrow(4)^2=a^2-(3)^2 \\
& \Rightarrow a^2=16+9=25
\end{aligned}
$$
Therefore, the required equation of ellipse is $\frac{x^2}{25}+\frac{y^2}{9}=1$.
19. Centre at $(0,0)$, major axis on the $y$-axis and passes through the points $(3,2)$ and $(1,6)$.
Ans. Since the major axis is along $y$-axis, therefore equation of ellipse is $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ And the ellipse passes through the point $(3,2)$ therefore $\frac{9}{b^2}+\frac{4}{a^2}=1$. $\qquad$
And the ellipse passes through the point $(1,6)$ therefore $\frac{1}{b^2}+\frac{36}{a^2}=1$.
Solving eq. (i) and (ii), we have $a^2=40, b^2=10$
Therefore, the required equation of ellipse is $\frac{x^2}{10}+\frac{y^2}{40}=1$.
20. Major axis on the $x$-axis and passes through the points $(4,3)$ and $(6,2)$.
Ans. Since the major axis is along $x$-axis, therefore equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ And the ellipse passes through the point $(4,3)$ therefore $\frac{16}{a^2}+\frac{9}{b^2}=1$.
And the ellipse passes through the point $(6,2)$ therefore $\frac{36}{a^2}+\frac{4}{b^2}=1$.
Solving eq. (i) and (ii), we have $a^2=52, b^2=13$
Therefore, the required equation of ellipse is $\frac{x^2}{52}+\frac{y^2}{13}=1$.
