Question

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$ be an ellipse such that LR = 10 and its eccentricity is equal to maximum value of quadratic expression f(t) = (5/12 )+ t—t² then (a² + b²) =

Answer 1

It has given that x²/a² + y²/b² = 1 be an ellipse such that LR = 10 and its eccentricity is equal to maximum value of quadratic experience, f(t) = (5/12) + t - t²

we have to find the value of (a² + b²)

solution : Latus rectum of an ellipse of equation x²/a² + y²/b² = 1, is given by 2b²/a

so LR = 2b²/a = 10

⇒b² = 5a .......(1)

f(t) = (5/12) + t - t²

differentiating w.r.t t we get,

f'(t) = 1 - 2t

at f'(t) = 0, t = 1/2

now again differentiating w.r.t t we get,

f"(t) = -2 < 0 hence at t = 1/2 we get maximum value of f(t)

so, maximum value of f(t) = 5/12 + 1/2 - 1/4

= 5/12 + 1/4

= 8/12

= 2/3

so eccentricity, e = maximum value of f(t) = 2/3

now using formula, b² = a²(1 - e²)

⇒b² = a²(1 - 2²/3²) = a²(1 - 4/9)

⇒5a = a²(5/9) [ from equation (1) ]

⇒a = 9

and b² = 5a = 45

now a² + b² = 9² + 45 = 81 + 45 = 126

Therefore the value of (a² + b²) is 126