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Eccentricity from Foci and Foot of Perpendicular | JEE

JEE Maths question with a full step-by-step solution.

Question
Let S(3,4)S(3,4) and S′(9,12)S'(9,12) be the foci of an ellipse and the foot of the perpendicular from SS to a tangent to the ellipse is (1,−4)(1,-4). Then the eccentricity of the ellipse is:
A313\dfrac{3}{13}
B413\dfrac{4}{13}
C513\dfrac{5}{13}correct
D713\dfrac{7}{13}
Solution
Step 1: The foot of the perpendicular from a focus to any tangent lies on the auxiliary circle, whose centre is the centre of the ellipse and radius is aa. Step 2: The centre is the midpoint of the foci.
C=(3+92,4+122)=(6,8).C=\left(\frac{3+9}{2},\frac{4+12}{2}\right)=(6,8).
 Step 3: Since (1,−4)(1,-4) lies on the auxiliary circle,
a=CP=(6−1)2+(8+4)2=25+144=169=13.a=CP=\sqrt{(6-1)^2+(8+4)^2}=\sqrt{25+144}=\sqrt{169}=13.
 Step 4: The distance between the foci gives aeae.
SS′=2ae=(9−3)2+(12−4)2=36+64=10  ⇒  ae=5.SS'=2ae=\sqrt{(9-3)^2+(12-4)^2}=\sqrt{36+64}=10 \;\Rightarrow\; ae=5.
 Step 5: Therefore
e=aea=513.e=\frac{ae}{a}=\frac{5}{13}.
 Correct answer: (3)
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