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, $12=16\left(1- e ^{2}\right)$
$\Rightarrow e =1 / 2$