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. $5x^3+3y^2-48=0$
• 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 C $5x^2+3y^2-32=0$

We know that equation of ellipse is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ …….(1)

Given that

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

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

Taking square both side and solving , we get

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

$a^2=\frac{5b^2}{3}$ …….(2)

$\because$ ellipse pass through (-3,1)

Then $x=-3,y=1$

Put in equation (1) we get

$\frac{{(-3)}^2}{a^2}+\frac{1^2}{b^2}=1$

$a^2+9b^2=a^2{b}^2$

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

$b^2=\frac{32}{5}$ (From equation (1) and (2) )

Put in equation (2) , we get $a^2=\frac{32}{3}$

the value of a and b put in equation (1), we get

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

$3x^2+5y^2=32$

This is required equation