JEE Main 2024 (Online) 30th January Evening ShiftQuestion 1
Let $A(\alpha,0)$ 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
Ellipse
Solution
Formulas used. (1) Section formula: $P=\frac{3A+7B}{10}$ for ratio $7:3$(2) Focus $(ae,0)$, directrix $x=\frac ae$, $b^2=a^2-a^2e^2$(3) Latus rectum $=\frac{2b^2}{a}$Step 1 - A and B. $5\alpha=50\Rightarrow A(10,0)$; $7\beta=50\Rightarrow B\left(0,\frac{50}{7}\right)$. Step 2 - Use (1). $P=\left(\frac{30}{10},\frac{7}{10}\cdot\frac{50}{7}\right)=(3,5)$. Step 3 - Focus. The perpendicular from $S$ to the $x$-axis is the line $x=ae$; it passes through $P$, so $ae=3$.Step 4 - Use (2). $\frac ae=\frac{25}{3}$. Multiply: $a^2=3\cdot\frac{25}{3}=25$, $b^2=25-9=16$. Step 5 - Use (3). $\frac{2\cdot16}{5}=\frac{32}{5}$. Answer: C