Let the length of a latus rectum of an ellipse be 10. If its eccentricity is the value of the function , then is equal to: [2025]

Question

Let the length of a latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ be 10. If its eccentricity is the minimum value of the function $f (t) = t^{2} + t + \frac{11}{12}, t \in R$ , then $a^{2} + b^{2}$ is equal to: [2025]

Smart Achievers · Accepted Answer

(1)

Length of latus rectum $= \frac{2 b^{2}}{a} = 10$ [Given]

$\Rightarrow 5 a = b^{2}$ ... (i)

Now, eccentricity is minimum value of

$f (t) = t^{2} + t + \frac{11}{12}$

$\Rightarrow f' (t) = 2 t + 1$

For critical point $f' (t) = 0$

$\Rightarrow 2 t + 1 = 0 \Rightarrow t = \frac{- 1}{2}$

Since, $f'' (t) = 2 > 0$ , so at $t = \frac{- 1}{2}$ , f(t) will give the minimum value.

$∴ f (\frac{- 1}{2}) = \frac{1}{4} - \frac{1}{2} + \frac{11}{12} = \frac{3 - 6 + 11}{12} = \frac{2}{3}$

$\Rightarrow e = \frac{2}{3} \Rightarrow \frac{2}{3} = \sqrt{1 - \frac{b^{2}}{a^{2}}}$

$\Rightarrow \frac{4}{9} = 1 - \frac{5 a}{a^{2}}$ [Using (i)]

$\Rightarrow \frac{4}{9} = 1 - \frac{5}{a} \Rightarrow \frac{5}{a} = 1 - \frac{4}{9} = \frac{5}{9} \Rightarrow a = 9$

$\Rightarrow b = \sqrt{45} = 3 \sqrt{5}$

$∴ a^{2} + b^{2} = {(9)}^{2} + {(3 \sqrt{5})}^{2} = 81 + 45 = 126$ .

Solution

Q.
Let the length of a latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ be 10. If its eccentricity is the minimum value of the function $f (t) = t^{2} + t + \frac{11}{12}, t \in R$ , then $a^{2} + b^{2}$ is equal to: [2025]

1 126

2 120

3 115

4 125

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