Question

Let \(E_1 : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,\ a \gt b\). Let \(E_2\) be another ellipse such that it touches the end points of major axis of \(E_1\) and the foci of \(E_2\) are the end points of minor axis of \(E_1\). If \(E_1\) and \(E_2\) have same eccentricities, then its value is :

• A. \(\frac{-1 + \sqrt{5}}{2}\)
• B. \(\frac{-1 + \sqrt{8}}{2}\)
• C. \(\frac{-1 + \sqrt{3}}{2}\)
• D. \(\frac{-1 + \sqrt{6}}{2}\)

Hint:

E2 has its major axis along the y-axis with semi-minor axis a and focal distance b; equate the two eccentricities.

Answer 1

The Correct Option is A

The end points of the major axis of \(E_1\) are \((\pm a, 0)\) and of the minor axis \((0, \pm b)\).
\(E_2\) has foci \((0, \pm b)\), so its major axis is along the y-axis with \(ae_2 = b\) (focal distance). It passes through \((\pm a, 0)\), so its semi-minor axis is a and its semi-major axis \(B\) satisfies \(B^2 = a^2 + b^2\).
\(e_2^2 = \frac{b^2}{a^2 + b^2}\) and \(e_1^2 = 1 - \frac{b^2}{a^2}\).
Let \(t = \frac{b^2}{a^2}\). \(e_1 = e_2\) gives \(1 - t = \frac{t}{1 + t} \Rightarrow 1 - t^2 = t \Rightarrow t^2 + t - 1 = 0\), so \(t = \frac{-1 + \sqrt{5}}{2}\) (positive root).
So the required value \(\frac{b^2}{a^2} = \frac{-1 + \sqrt{5}}{2}\).
Answer: (1) \(\frac{-1 + \sqrt{5}}{2}\)