Question

Let $S(3,4)$ and $S^{\prime }(9,12)$ 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

The correct option is C $\frac{5}{13}$

$S{S}^{\prime }=2ae$, where $a$ and $e$ are length of semi-major axis and eccentricity respectively.
$\therefore \sqrt{(9-3{)}^2+(12-4{)}^2}=2ae$
$\Rightarrow ae=5$
Centre is mid-point of $S{S}^{\prime }$
$\therefore$ Centre $\equiv (6,8)$
Let the equation of the 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.
So, $(1,-4)$ lies on auxiliary circle, i.e., $(1-6{)}^2+(-4-8{)}^2=a^2$
$\Rightarrow a=13$
Since, $ae=5$ $\Rightarrow e=\frac{5}{13}$