Question

$S(3,4)$ and $S{}^{\prime }(9,12)$ are the foci of an ellipse and the foot of the perpendicular from $S$ to a tangent to the ellipse is $(1,-4)$. Then the eccentricity of the ellipse is:

• A. $\frac{3}{13}$
• B. $\frac{4}{13}$
• C. $\frac{5}{13}$
• D. $\frac{7}{13}$

Answer 1

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

Given foci are $S(3,4)$ and $S^{\prime }(9,12)$ and foot of the perpendicular from $S$ to a tangent to the ellipse is $(1,-4)$.

We know that distance between foci is $2ae=\sqrt{(9-3{)}^2+(12-4{)}^2}=\sqrt{36+64}=10$.........(1), where $e$ is eccentricity of ellipse.
$2a$ is the length of major axis of ellipse. Center of ellipse is $(6,8)$.

We know that distance between center and foot of perpendicular to tangent is $CP=a⟹a=\sqrt{(6-1{)}^2+(8-(-4){)}^2}=\sqrt{25+144}=13$.

From(1), we get eccentricity $e=\frac{5}{13}$.