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

Answer: (D)

Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
It passes through $(-3,1)$
so $\frac{9}{a^2}+\frac{1}{b^2}=1(i)$
Also, $b^2=a^2(1-2/5)$
$\Rightarrow 5b^2=3a^2....(ii)$
Solving (i) and (ii) we get $a^2-\frac{32}{3},b^2=\frac{32}{5}$
So the equation of the ellipse is $3x^2+5v^2=32$