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Question

# Let S(3,4) and S′(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
45
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B
57
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C
513
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D
713
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Solution

## The correct option is C 513SS′=2ae, where a and e are length of semi-major axis and eccentricity respectively. ∴√(9−3)2+(12−4)2=2ae ⇒ae=5 Centre is mid-point of SS′ ∴ Centre ≡(6,8) Let the equation of the auxiliary circle be (x−6)2+(y−8)2=a2 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=a2 ⇒a=13 Since, ae=5 ⇒e=513

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