JEE Mathematics 2024 Question with Solution

Given: $A(10,0)$ and $B(0,\beta)$ lie on $5x+7y=50$. Point $P$ divides $AB$ in the ratio $7:3$. The directrix of the ellipse is $x=\frac{25}{3}$.

Find: the length of the latus rectum.

First find $\beta$ using the line equation:

So,

Hence,

Using the section formula for internal division in the ratio $7:3$,

Therefore,

For the ellipse

the focus is $S=(ae,0)$. Since a perpendicular from $S$ passes through $P(3,5)$, the focus has $x$-coordinate $3$. Hence,

The directrix of the ellipse is

So,

Using $ae=3$ and $\frac{a}{e}=\frac{25}{3}$, we get

Now use

Thus,

The length of the latus rectum of the ellipse is

Therefore,

So, the length of the latus rectum is $\frac{32}{5}$. The correct option is D.

Given: $A(10,0)$, $B(0,\beta)$, line $5x+7y=50$, ratio $7:3$, directrix $x=\frac{25}{3}$.

Find: the latus rectum length of the ellipse.

Since $B(0,\beta)$ lies on $5x+7y=50$,

Hence,

Now divide $AB$ in the ratio $7:3$:

So the perpendicular from the focus passes through $P$ means the focus is vertically aligned with $P$, giving

Thus,

For the standard ellipse, the directrix corresponding to the right focus is

Given directrix:

Combining with

we obtain

Therefore,

and then

Now,

where

Thus,

Hence latus rectum length is

Therefore, the required answer is $\frac{32}{5}$.