Ellipse
Directrix, Focus and Latus Rectum
nta_pyq_2024_jan
Grade 11
Question:
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 $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ 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
$\dfrac{25}{3}$
$\dfrac{32}{9}$
$\dfrac{25}{9}$
$\dfrac{32}{5}$
Step-by-Step Solution
Key Concept: Find $A$ and $B$ from the line. Use section formula to find $P$. Since perpendicular from $S$ to $x$-axis passes through $P$, the focus $S$ has the same $x$-coordinate as $P$. Use directrix $x=25/3$ and focus $S=(ae,0)$ with $a/e=25/3$ to find $a$ and $b$.
$A=(10,0)$, $B=(0,50/7)$, $P=(3,5)$. $ae=3$, $a/e=25/3\Rightarrow a=5,e=3/5$. $b^2=16$. LR$=\frac{2\times16}{5}=\frac{32}{5}$.
Correct Answer: 4