Question

Let $A(10, 0)$ and $B(0, \beta)$ be the points on the line $5x + 7y = 50$. Let the point $P$ divide the line segment $AB$ internally in the ratio $7 : 3$. Let $3x - 25 = 0$ be a directrix of the ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and the corresponding focus be $S$. If from $S$, the perpendicular on the x-axis passes through $P$, then the length of the latus rectum of $E$ is equal to

• A. $\frac{25}{3}$
• B. $\frac{32}{9}$
• C. $\frac{32}{5}$
• D. $\frac{25}{9}$

Answer 1

Answer: (C)

$\frac{32}{5}$

Answer 2

Solution: Substitute $x = 0$ and $y = \beta$ in the line equation $5x + 7y = 50$ to find $\beta$:

$7\beta = 50 \Rightarrow \beta = \frac{50}{7}.$

Thus, $B = \left(0, \frac{50}{7}\right).$

Using the section formula, $P = (3, 5)$, which divides $AB$ in the ratio $7 : 3$.

The directrix is $x = \frac{25}{3}$, so $a = \frac{25}{3}$ and $e = \frac{3a}{25}$. Given that $ae = 3$, solving yields $a = 5$ and $b = 4$.

The length of the latus rectum $LR$ is:

$LR = \frac{2b^2}{a} = \frac{32}{5}.$