QJEEM20Ma4386: JEE Main 2020 Mathematics - Class 11 JEEM20Ma4386 Solution: Let (a > b) be a given ellipse, length of whose latus rectum is …

Question

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Let

{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1

(a > b) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function,

\phi \left( t \right) = {5 \over {12}} + t - {t^2}

, then a² + b² is equal to :

Accepted Answer

Given ellipse

{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1

(a > b)

Length of latus rectum

= {{2{b^2}} \over a} = 10

\phi (t) = {5 \over {12}} + t - {t^2}

= {8 \over {12}} - {\left( {t - {1 \over 2}} \right)^2}

\therefore

\phi {(t)_{\max }} = {8 \over {12}} = {2 \over 3}

\therefore

eccentricity (e) =

{2 \over 3}

Also,

{e^2} = 1 - {{{b^2}} \over {{a^2}}}

\Rightarrow {4 \over 9} = 1 - {{{b^2}} \over {{a^2}}}

\Rightarrow {{{b^2}} \over {{a^2}}} = {5 \over 9}

\Rightarrow {{{b^2}} \over a} \times {1 \over a} = {5 \over 9}

\Rightarrow {5 \over a} = {5 \over 9}

\Rightarrow a = 9

\therefore

{b^2} = 5 \times 9 = 45

\therefore

{a^2} + {b^2} = 81 + 45 = 126

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