Question

The ellipse $E_1;\frac{x^2}{9}+\frac{y^2}{4}=1$ is incribed in a rectangle R whose sides are parallel to the coordinates axis. 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 $\frac{1}{2}$

REF.image

For $E_1,a=3,b=2$

$\to$ vertex = (a,0),(0,b) = (3,0),(0,2)

$\Rightarrow P(3,2)$

For $E_2,b=4\{vertex=(6,4\left)\right\}$

and it passes through(3,2)

$\Rightarrow \frac{3^2}{a^2}+\frac{2^2}{b^2}=1\Rightarrow \frac{9}{a^2}=1-\frac{4}{16}=\frac{3}{4}\Rightarrow a^2=\frac{36}{3}\Rightarrow a=2\sqrt{3}$

As a comes less than b, $\Rightarrow$ Ellipse axis is along y

$\Rightarrow e^2=1-\frac{a^2}{b^2}=1-\frac{12}{16}=1-\frac{3}{4}=\frac{1}{4}\Rightarrow e=\frac{1}{2}\Rightarrow \left(C\right)$