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Question

# S(3, 4) and S’(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

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Solution

## The correct option is C SS’ = 2ae ⇒ 2ae = 10 ⇒ ae = 5 Centre C is the midpoint of SS’. ∴ C = (6, 8) Equation of the auxiliary circle is (x−6)2+(y−8)2=a2 Since the locus of the foot of the perpendicular from locus to a tangent of the ellipse is called the auxiliary circle, we have (1, –4) lies on the auxiliary circle. ∴ a = 13. Hence e=513

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