Question

Let from the point with abscissa $25$, two tangents be drawn to the ellipse $24x^2+25y^2=600$ with foci at $S_1$ and $S_2$. The points of contact of tangents are $A$ and $B$. Let the distance of $A$ from $S_1$ be $\frac{60}{13}$ units and $p$ be the sum of distance of $A$ from the directrix corresponding to $S_2$ and the distance of $B$ from $S_2.$ Then the value of $13p$ is

Answer 1

$\frac{x^2}{25}+\frac{y^2}{24}=1$
$a^2=25,b^2=24$
Now, $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{24}{25}}=\frac{1}{5}$
Directrix is $x=\pm \frac{a}{e}=\pm 25$
Clearly, the point with abscissa $25$ lies on the directrix $x=25$
So, $AB$ is a focal chord
and $\frac{1}{SA}+\frac{1}{SB}=\frac{2a}{b^2}$
$\Rightarrow \frac{1}{SA}+\frac{1}{SB}=\frac{5}{12}$
As given $S_{1}A=\frac{60}{13},$ then $S_{1}B=5\dots \left(1\right)$
Also, $S_{1}A+S_{2}A=2a=10\dots \left(2\right)$
and $S_{1}B+S_{2}B=10\dots \left(3\right)$

From equations $\left(1\right),\left(2\right)$ and $\left(3\right),$
$S_{2}A=\frac{70}{13},S_{2}B=5$
$\text{Distance of}A\text{from directrix corresponding to}{S}_2$
$=\frac{1}{e}\times \text{Distance from focus}$
$=5\times \frac{70}{13}=\frac{350}{13}$

Now, $p=\frac{350}{13}+S_{2}B=\frac{350}{13}+5=\frac{415}{13}$