Let the length of the latus rectum of an ellipse , be 30. If its eccentricity is the maximum value of the function then is equal to [2026]

Question

Let the length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ , be 30. If its eccentricity is the maximum value of the function $f (t) = - \frac{3}{4} + 2 t - t^{2},$ then $(a^{2} + b^{2})$ is equal to [2026]

Smart Achievers · Accepted Answer

(4)

$f (t) = \frac{- 3}{4} + 2 t - t^{2}$

$f (t) |_{maximum} = \frac{1}{4} = e \Rightarrow e^{2} = \frac{1}{16} \Rightarrow \frac{a^{2} - b^{2}}{a^{2}} = \frac{1}{16} . . . (1)$

$∵$ $\frac{2 b^{2}}{a} = 30 \Rightarrow b^{2} = 15 a . . . (2)$

$By (1) & (2)$

$16 (a^{2} - 15 a) = a^{2} \Rightarrow 15 a^{2} - 16 \times 15 a = 0$

$a = 16$

$b^{2} = 240$

$a^{2} + b^{2} = 256 + 240$

$= 496$

Solution

Q.
Let the length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ , be 30. If its eccentricity is the maximum value of the function $f (t) = - \frac{3}{4} + 2 t - t^{2},$ then $(a^{2} + b^{2})$ is equal to [2026]

1 256

2 516

3 276

4 496

Want Us to Email you About Special Offers & Updates?