Question

Equation of the ellipse whose axes are the axes of coordinates and which passes through the point $(-3,1)$ and has eccentricity $\sqrt{\frac{2}{5}}$ is

• A. $5x^2+3y^2–48=0$
• B. $3x^2+5y^2–15=0$
• C. $5x^2+3y^2–32=0$
• D. $3x^2+5y^2–32=0$

Answer 1

The correct option is D

$3x^2+5y^2–32=0$

Step 1. Find the equation of the ellipse;

Let the ellipse equation be

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

and $e=\sqrt{\frac{2}{5}}$

or, $\sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{\frac{2}{5}}$

Squaring both sides,

$5(a^2-b^2)=2a^2$

or, $a^2=\frac{5}{3}{b}^2...\left(\mathrm{i}\right)$

Also, the ellipse passes through the point $(-3,1)$

$\Rightarrow$$\frac{9}{a^2}+\frac{1}{b^2}=1$

$\Rightarrow$ $9b^2+a^2=a^2{b}^2...\left(\mathrm{ii}\right)$

Step 2. Find the value of $b^2$ and $a^2$by using the obtained expression:

from equation $\left(\mathrm{i}\right)$ put the value of $a^2$ in equation $\left(\mathrm{ii}\right)$

$\Rightarrow$$9b^2+\frac{5}{3}{b}^2=\frac{5}{3}{b}^4$

$\Rightarrow$ $\frac{32}{3}=\frac{5}{3}{b}^2$

$\Rightarrow$ $b^2=\frac{32}{5}$

Put the value of $b^2$ in $a^2$:

$\Rightarrow$ $a^2=\frac{32}{3}$

$\therefore$The Equation of the ellipse will be:

$\frac{3x^2}{32}+\frac{5y^2}{32}=1$

$\Rightarrow$$\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{5}{\mathit{y}}^{\mathbf{2}}\mathbf{-}\mathbf{32}\mathbf{=}\mathbf{0}$

Hence, Option ‘D’ is Correct.