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Q.

Let x2a2+y2b2=1(a>b) be a given ellipse, length of whose latus rectum is 10 . If its eccentricity is  the maximum value of the function, f(t)=512+t-t2, then a2+b2 is equal to :

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a

145

b

126

c

116

d

135

answer is B.

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Detailed Solution

10=2b2a⇒b2=5a  f(t)=23-t-122⇒C=23b2=a21-c2⇒5a=a21-49⇒a=9;b2=45⇒a2+b2=81+45=126
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