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

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

In this situation, as ellipse circumscribes the rectangle, then it must pass through all four vertices. Let The equation of an ellipse $E_2$ be
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where a < b and b=4

Also it passes through (3,2).
$\Rightarrow \frac{9}{a^2}+\frac{4}{b^2}=1\Rightarrow \frac{9}{a^2}+\frac{1}{4}=1or{a}^2=12$
Eccentricity of $E_2,=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{12}{16}}=\sqrt{\frac{1}{4}}e=\frac{1}{2}$