Question

Let $S \equiv(3,4)$ and $S ^{\prime} \equiv(9,12)$ be two foci of an ellipse. If the coordinates of the foot of the perpendicular from focus $S$ to a tangent of the ellipse is $(1,-4)$ then the eccentricity of the ellipse is

• A. $\frac{4}{5}$
• B. $\frac{5}{7}$
• C. $\frac{5}{13}$
• D. $\frac{7}{13}$

Answer 1

Answer: (C)

$SS ^{\prime}=2 ae$, where a and e are length of semi-major axis and eccentricity respectively.
$\therefore \sqrt{(9-3)^2+(12-4)^2}=2 a e $
$\therefore \text { ae }=5$
$\therefore \text { centre is mid-point of SS' }$
$\therefore \text { centre } \equiv(6,8)$
Let the equation of auxiliary circle be $(x-6)^2+(y-8)^2=a^2$
We know that the foot of the perpendicular from the focus on any tangent lies on the auxiliary circle
$\therefore (1,-4) \text { lies on auxiliary circle } $
$\text { i.e. } (1-6)^2+(-4-8)^2=a^2 \Rightarrow a =13$
$\because \text { ae }=5 \Rightarrow e =5 / 13$