Let $E_{1} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b .$ 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

\frac{- 1 + \sqrt{6}}{2}

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\frac{- 1 + \sqrt{8}}{2}

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\frac{- 1 + \sqrt{3}}{2}

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\frac{- 1 + \sqrt{5}}{2}

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Solution

The correct option is D $\frac{- 1 + \sqrt{5}}{2}$
Given: $E_{1} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b$
Let the other ellipse be
$E_{2} = \frac{x^{2}}{A^{2}} + \frac{y^{2}}{B^{2}} = 1 (B > A)$
Focus $= (0, B e)$
So,
$(0, B e) = (0, b) \Rightarrow e = \frac{b}{B} \dots (1)$
Also,
$(A, 0) = (a, 0) \Rightarrow A = a \dots (2)$
The eccentricity is given same, so
$1 - \frac{b^{2}}{a^{2}} = 1 - \frac{A^{2}}{B^{2}} \Rightarrow \frac{b^{2}}{a^{2}} = \frac{a^{2} e^{2}}{b^{2}} \Rightarrow e^{2} = \frac{b^{4}}{a^{4}} \Rightarrow e = \frac{b^{2}}{a^{2}} = 1 - e^{2} \Rightarrow e^{2} + e - 1 = 0 \Rightarrow e = \frac{- 1 \pm \sqrt{5}}{2} ∴ e = \frac{- 1 + \sqrt{5}}{2}$

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and

E_{2} = \frac{x^{2}}{k^{2}} + \frac{y^{2}}{b^{2}} - 1 = 0, (k < b) E_{2}

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E_{1}

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E_{1} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - 1 = 0, (a > b)

and

E_{2} : \frac{x^{2}}{k^{2}} + \frac{y^{2}}{b^{2}} - 1 = 0, (k < b)

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E_{1}

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E_{2} = p \times L L R of E_{1},

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p =

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