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Let S=(3,4) and S'=(9,12) be two foci of...

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

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The correct Answer is:
`5//13`
Centre of ellipse `C-=((3+9)/(2),(4+12)/(2))-=(6,8)`
Food of perpendicular from focus upon any tangent lies on the auxiliary circle.
Thus, P(1,-4) lies on the auxiliary circle.
So, CP=13=a
Also, distance between foci,
`SS'sqrt((3-9)^(2)+(4-12)^(2))=10=2ae`
`:. e=(5)/(13)`
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