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)

$\frac{1}{2}$

Answer 2

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}