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JEE Advanced · Mathematics · 14. Ellipse

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

  1. A \(\frac{\sqrt{2}}{2}\)
  2. B \(\frac{\sqrt{3}}{2}\)
  3. C \(\frac{1}{2}\)
  4. D \(\frac{3}{4}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{2}\)

Step-by-step Solution

Detailed explanation

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 \(A(3,2)\)

\(\therefore \quad \frac{9}{a^{2}}+\frac{4}{16}=1 \Rightarrow a^{2}=12\)

\(\therefore \quad e=\sqrt{1-\frac{12}{16}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)
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