JEE MAIN - Mathematics (2020 - 4th September Morning Slot - No. 16)
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 :
$$\phi \left( t \right) = {5 \over {12}} + t - {t^2}$$, then a2 + b2 is equal to :
145
126
135
116
Explanation
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, $${e^2} = 1 - {{{b^2}} \over {{a^2}}}$$
$$ \Rightarrow {4 \over 9} = 1 - {{{b^2}} \over {{a^2}}}$$
$$ \Rightarrow {{{b^2}} \over {{a^2}}} = {5 \over 9}$$
$$ \Rightarrow {{{b^2}} \over a} \times {1 \over a} = {5 \over 9}$$
$$ \Rightarrow {5 \over a} = {5 \over 9}$$
$$ \Rightarrow a = 9$$
$$ \therefore $$ $${b^2} = 5 \times 9 = 45$$
$$ \therefore $$ $${a^2} + {b^2} = 81 + 45 = 126$$
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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