JEE MAIN - Mathematics (2021 - 22th July Evening Shift - No. 14)
Let $${E_1}:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1,a > b$$. Let E2 be another ellipse such that it touches the end points of major axis of E1 and the foci of E2 are the end points of minor axis of E1. If E1 and E2 have same eccentricities, then its value is :
$${{ - 1 + \sqrt 5 } \over 2}$$
$${{ - 1 + \sqrt 8 } \over 2}$$
$${{ - 1 + \sqrt 3 } \over 2}$$
$${{ - 1 + \sqrt 6 } \over 2}$$
Explanation
_22th_July_Evening_Shift_en_14_1.png)
$${e^2} = 1 - {{{b^2}} \over {{a^2}}}$$
$${e^2} = 1 - {{{a^2}} \over {{c^2}}}$$
$$ \Rightarrow {{{b^2}} \over {{a^2}}} = {{{a^2}} \over {{c^2}}}$$
$$ \Rightarrow {c^2} = {{{a^4}} \over {{b^2}}} \Rightarrow c = {{{a^2}} \over b}$$
Also b = ce
$$ \Rightarrow c = {b \over e}$$
$${b \over e} = {{{a^2}} \over b}$$
$$ \Rightarrow e = {{{b^2}} \over {{a^2}}} = 1 - {e^2}$$
$$ \Rightarrow {e^2} + e - 1 = 0$$
$$ \Rightarrow e = {{ - 1 + \sqrt 5 } \over 2}$$
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