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. $\frac{3}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$


Let the equation of $E_2$ be
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Clearly the point $P$ has coordinates $(3,2)$
So, $(3,2)$ and $(0,4)$ satisfy $E_2$
$\frac{16}{b^2}=1\Rightarrow b^2=16$
and $\frac{9}{a^2}+\frac{4}{b^2}=1$
$\frac{9}{a^2}=\frac{3}{4}\Rightarrow a^2=12$
$e=\sqrt{1-\frac{12}{16}}=\frac{1}{2}$