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The foci of an ellipse are S(3,1) and S'...

The foci of an ellipse are `S(3,1)` and `S'(11,5)`The normal at P is `x+2y-15=0` Then point P is

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The correct Answer is:
(17,1)

We know that normal at P bisects `angleSPS'`
Thus, the image of S w.r.t.x+2y-15=0 lies on S'P.
Now, image of S(3,1) w.r.t.x+2y-15 ius Q(7,9).
Equation of S'Q is x+y-16=-0
Clearly, from the figure, P is point of intersection of S'Q and normal, which is (17,-1)
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