Question

Two tangents are drawn from a point with abscissa 25, to the ellipse $24x^2+25y^2=600$ with foci at $S_1$ and $S_2$. The points of contact of the tangents are $A$ and $B.$ If the distance of $A$ from $S_1$ is $\frac{60}{13}$ units, then

• A. The distance of $A$ from $S_2$ is $\frac{80}{13}$ units
• B. The distance of $B$ from $S_1$ is 5 units
• C. The distance of $B$ from $S_2$ is 5 units
• D. The distance of $A$ from the directrix corresponding to $S_2$ is $\frac{350}{13}$ units.

Answer 1

Answer: (B), (C), (D)

The correct options are
B

The distance of $B$ from $S_1$ is 5 units

C

The distance of $B$ from $S_2$ is 5 units

D

The distance of $A$ from the directrix corresponding to $S_2$ is $\frac{350}{13}$ units.

Given ellipe: $\frac{x^2}{25}+\frac{y^2}{24}=1,a=5,b=2\sqrt{6}$

Eccentricity, $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{24}{25}}=\frac{1}{5}$

Equation of the directrix is $x=\frac{a}{e}$, i.e., $x=25$

So, the point from which the tangents are drawn, lie on the directrix.

Therefore, $AB$ is the focal chord.

$\Rightarrow \frac{1}{S_{1}A}+\frac{1}{S_{1}B}=\frac{2a}{b^2}$

$=\frac{13}{60}+\frac{1}{S_{1}B}=\frac{10}{24}$

$\Rightarrow S_{1}B=5$

Also $S_{1}A+S_{2}A=2a=10=S_{1}B+S_{2}B$

$\Rightarrow S_{2}A=\frac{70}{13},S_{2}B=5$

Distance of $A$ from the directrix corresponding to $S_2=\frac{1}{e}\times \text{distance from focus}=\frac{70}{13}\times 5=\frac{350}{13}$