Let [x^2 a^2 + y^2 b^2 = 1] (a b) be a given ellipse, length of whose latus rectum is 10…

Question

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

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

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

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

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

$$ \Rightarrow a = 9$$

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

$$ \therefore $$ ${a^2} + {b^2} = 81 + 45 = 126$

Solutions · 1