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### Question

#### If the tangent at Point P to the ellipse 16x^{2} + 11y^{2} = 256 is also the tangent to the circle x^{2} + y^{2} - 2x = 15, then the eccentric angle of point P is

#### The correct answer is:

#### The equation of tangent at point to the ellipse

16x^{2} + 11y^{2} = 256 is

16x (4cosq) + 11y = 256

4xcosq + y sinq = 16

This touches the circle

(x + 1)^{2} + y^{2} = 16

So, = 4

Þ (cosq - 4)^{2} = 11 + 5cos^{2}q

4cos^{2}q + 8cosq - 5 = 0

\ cosq =

\ q = ±

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