question_answer

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 ellipse \[{{E}_{2}}\] is

A) \[\frac{\sqrt{2}}{2}\]

B) \[\frac{\sqrt{3}}{2}\]

C) \[\frac{1}{2}\]

D) \[\frac{3}{4}\]

Correct Answer: C

Solution :

Let the ellipse \[{{E}_{2}}=\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] as  it is passing through \[\left( 0,4 \right){{b}^{2}}=16\]

Also it is passes through (3, 2).

\[{{a}^{2}}=12\]

\[\therefore \,\,\,e=\sqrt{1-\frac{{{a}^{2}}}{{{b}^{2}}}}=\sqrt{1-\frac{12}{16}}\]

\[e=\frac{1}{2}\]