The point(s) on the ellipse 16x2+11y2=256 where the common tangent to it and the circle x2+y2−2x=15 touch is
A
(2,8√311)
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B
(2,8√311)
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C
(−2,−8√311)
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D
(2,−8√311)
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Solution
The correct options are A(2,8√311) D(2,−8√311) Ellipse: x216+11y2162=1 Tangent to the ellipse at point P(4cosθ,16√11sinθ) is x4cosθ+√11y16sinθ=1 It also touches the circle x2+y2−2x=15(x−1)2+y2=42
∴distance of the tangent to the centre of the circle = radius ∴|14cosθ−1|√cos2θ16+11sin2θ256=4|cosθ−4|=√11+5cos2θcos2θ+16−8cosθ=11+5cos2θ4cos2θ+8cosθ−5=0(2cosθ−1)(2cosθ+5)=0cosθ=12⇒θ=±π3