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QJEEM21Maf48c: JEE Main 2021 Mathematics - Class 11 JEEM21Maf48c Solution: Let . Let E 2 be another ellipse such that it touches the end…

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 :

Multiple Choice Options

  • A −1+52{{ - 1 + \sqrt 5 } \over 2}
  • B −1+82{{ - 1 + \sqrt 8 } \over 2}
  • C −1+32{{ - 1 + \sqrt 3 } \over 2}
  • D −1+62{{ - 1 + \sqrt 6 } \over 2}

Step-by-Step Verified Solution

JEE Main 2021 (Online) 22th July Evening Shift Mathematics - Ellipse Question 52 English Explanation
e2=1−b2a2{e^2} = 1 - {{{b^2}} \over {{a^2}}}

e2=1−a2c2{e^2} = 1 - {{{a^2}} \over {{c^2}}}

⇒b2a2=a2c2\Rightarrow {{{b^2}} \over {{a^2}}} = {{{a^2}} \over {{c^2}}}

⇒c2=a4b2⇒c=a2b\Rightarrow {c^2} = {{{a^4}} \over {{b^2}}} \Rightarrow c = {{{a^2}} \over b}

Also b = ce

⇒c=be\Rightarrow c = {b \over e}

be=a2b{b \over e} = {{{a^2}} \over b}

⇒e=b2a2=1−e2\Rightarrow e = {{{b^2}} \over {{a^2}}} = 1 - {e^2}

⇒e2+e−1=0\Rightarrow {e^2} + e - 1 = 0

⇒e=−1+52\Rightarrow e = {{ - 1 + \sqrt 5 } \over 2}

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