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Question

The point(s) on the ellipse 16x2+11y2=256
where the common tangent to it and the circle x2+y22x=15 touch is

A
(2,8311)
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B
(2,8311)
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C
(2,8311)
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D
(2,8311)
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Solution

The correct options are
A (2,8311)
D (2,8311)
Ellipse:
x216+11y2162=1
Tangent to the ellipse at point
P(4cosθ,1611sinθ) is
x4cosθ+11y16sinθ=1
It also touches the circle x2+y22x=15(x1)2+y2=42

distance of the tangent to the centre of the circle = radius
|14cosθ1|cos2θ16+11sin2θ256=4|cosθ4|=11+5cos2θcos2θ+168cosθ=11+5cos2θ4cos2θ+8cosθ5=0(2cosθ1)(2cosθ+5)=0cosθ=12θ=±π3

Hence, the points P=(2,±8311)

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