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)

$\frac{1}{2}$

The ellipse $E_1$ touches $x$ axis at $(\pm 3,0)$ and $y$ axis at $(0,\pm 2)$

Since the ellipse is inscribed in rectangle $R$ whose sides are parallel to the coordinate axis , the vertices of rectangle are $(\pm 3,\pm 2)$

Let the equation of ellipse $E_2$ be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

The ellipse circumscribes the rectangle $R$ , so the vertices of rectangle lies on ellipse $E_2$

Therefore we get $\frac{9}{a^2}+\frac{4}{b^2}=1$

Given that ellipse $E_2$ passes through$(0,4)$

So we get $b=4$ and $a^2=12$

We know that $a^2=b^2(1-e^2)$

$\Rightarrow e^2=\frac{1}{4}$

$\Rightarrow e=\frac{1}{2}$