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 B $\frac{\sqrt{3}}{2}$

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 $\frac{9}{a^2}+\frac{4}{16}=1$

$\frac{9}{a^2}+\frac{1}{4}=1$

$\frac{9}{a^2}=1-\frac{1}{4}$

$\frac{9}{a^2}=\frac{3}{4}$

$\frac{3}{a^2}=\frac{1}{4}$

$a^2=12$

So,$12=16(1-e^2)$

or $-16e^2=12-16=-4$

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

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