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)

$S{S}^{\prime }=2ae$, where a and e are length of semi-major axis and eccentricity respectively.
$\therefore \sqrt{(9-3{)}^2+(12-4{)}^2}=2ae$
$\therefore \text{ ae }=5$
$\therefore \text{ centre is mid-point of SS’ }$
$\therefore \text{ centre }\equiv (6,8)$
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
$\therefore (1,-4)\text{ lies on auxiliary circle }$
$\text{ i.e. }(1-6{)}^2+(-4-8{)}^2=a^2\Rightarrow a=13$
$\because \text{ ae }=5\Rightarrow e=5/13$