Question

The ellipse $E_1:\frac{x^2}{9}+\frac{y^2}{4}=1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_2$ passing through the point (0,4) circumscribes the rectangle $R$. The eccentricity of the ellipse $E_2$ is

• A. $\frac{\sqrt{2}}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{2}$
• D. $\frac{3}{4}$

Answer 1

Answer: (C)

Equation of ellipse is $(y+2)(y-2)+\lambda (x+3)(x-3)=0$
It passes through $(0,4)$
$\Rightarrow \lambda =\frac{4}{3}$
Equation of ellipse is $\frac{x^2}{12}+\frac{y^2}{16}=1$
$e=\frac{1}{2}$
Alternate Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ as it is passing through (0,4) and (3,2) .
So, $b^2=16$ and $\frac{9}{a^2}+\frac{4}{16}=1$
$\Rightarrow a^2=12$
So, $12=16\left(1-e^2\right)$
$\Rightarrow e=1/2$