Question

If S(3, 4) & S’(9, 12) are two foci of an ellipse. If the foot of perpendicular from S on a tangent to the ellipse has the coordinates (1, –4), then eccentricity of the ellipse is

• A. $\frac{5}{13}$
• B. $\frac{7}{13}$
• C. $\frac{1}{5}$
• D. $\frac{1}{7}$

Answer 1

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

Locus of foot of perpendiculars drawn from foci on any tangent to the ellipse is its auxiliary circle. i.e.,
(1, -4) lies on auxiliary circle.
Now center of ellipse = $(\frac{3+9}{2},\frac{4+12}{2})=(6,8)$
Distance between foci $=2ae=10\Rightarrow ae=5$
Equation of auxiliary circle $\Rightarrow (x–6{)}^2+(y-8{)}^2=a^2$
Passes through (1, –4) $\Rightarrow 25+144=a^2\Rightarrow a=13$
$\therefore ae=5\Rightarrow e=\frac{5}{13}$