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

  $\frac{\sqrt{3}}{2}$
• Correct answer
  C

  $\frac{1}{2}$
• D

  $\frac{3}{4}$

Solution

As rectangle $A B C D$ circumscribed the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1, \quad \therefore A=(3,2)$
Let the ellipse circumscribing the rectangle $A B C D$ is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1...(i)$ Given that ellipse (i) passes through $(a, 4)$ $\therefore b^{2}=16$ Also ellipse (i) passes through $A(3,2)$ $\therefore \quad \frac{9}{a^{2}}+\frac{4}{16}=1 \Rightarrow a^{2}=12$ $\therefore \quad e=\sqrt{1-\frac{12}{16}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

Hence, the correct answer is option: 'C'.