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 :

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

Answer: B. 126 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}} ight)^2} 	herefore \phi {(t)_{\max }} = {8 \over {12}} = {2 \over 3} 	herefore 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} 	imes {1 \over a} = {5 \over 9} \Rightarrow {5 \over a} = {5 \over 9} \Rightarrow a = 9 	herefore {b^2} = 5 	imes 9 = 45 	herefore {a^2} + {b^2} = 81 + 45 = 126

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 :

Answer

Explanation

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