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

PLAN Equation of an ellipse is
$\hspace10mm \frac{x^2}{a^2} + \frac {y^2}{b^2} = 1 \hspace10mm \because [a > b]$
Eccentricity, $\hspace15mm{e^2} = 1 - \frac{b^2}{a^2}\hspace10mm \because [a > b]$
D escription Situation As ellipse circumscribes the rectangle, then it must pass through all four vertices.
Let the equation of an ellipse $E_2$ be
$\hspace 15mm \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$ where a < b and b = 4. $ $
$\frac{x^2}{a^2}+\frac{x^2}{a^2}=1$, where a < b and 6=4.
${\implies}\hspace15mm \frac{9}{a^2} + \frac{4}{b^2} = 1\hspace12mm [\because b=4]$
${\implies}\hspace15mm \frac{9}{a^2} + \frac{1}{4} = 1\, or\, a^2 = 12$
Eccentricity of $E_2$,$\hspace6mm e^2 - 1 - \frac{a^2}{b^2}= 1 - \frac {12}{16} = \frac {1}{4} \hspace3mm [\because a< b ]$
$\therefore \hspace13mm e = \frac{1}{2}$