JEE Mathematics 2025 Question with Solution

Given: The foci of the ellipse are $(2, 5)$ and $(2, -3)$, and the eccentricity is $e = \frac{4}{5}$.

Find: The length of the latus-rectum.

The distance between the foci is

$$2c = |5 - (-3)| = 8$$

So,

$$c = 4$$

Using the relation

$$e = \frac{c}{a}$$

we get

$$\frac{4}{5} = \frac{4}{a}$$

Hence,

$$a = 5$$

Now,

$$b^2 = a^2 - c^2 = 25 - 16 = 9$$

The length of the latus-rectum of an ellipse is

$$\frac{2b^2}{a}$$

Therefore,

$$\text{Length of latus-rectum} = \frac{2 \cdot 9}{5} = \frac{18}{5}$$

Therefore, the correct option is D.

Given: The foci are $(2,5)$ and $(2,-3)$, with eccentricity $\frac{4}{5}$.

Find: The length of the latus-rectum.

The center is the midpoint of the foci:

$$\left(\frac{2+2}{2}, \frac{5+(-3)}{2}\right) = (2,1)$$

The distance from the center to either focus is

$$c = \sqrt{(2-2)^2 + (5-1)^2} = 4$$

Using

$$e = \frac{c}{a}$$

we have

$$\frac{4}{5} = \frac{4}{a}$$

So,

$$a = 5$$

Now use

$$b^2 = a^2 - c^2$$

Thus,

$$b^2 = 25 - 16 = 9$$

Finally,

$$\text{LR} = \frac{2b^2}{a} = \frac{2 \cdot 9}{5} = \frac{18}{5}$$

Therefore, the length of the latus-rectum is $\frac{18}{5}$, so the correct option is D.