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Question

The normal at a point P on the ellipse x2+4y2=16 intersects the x-axis at Q. If M is the midpoint of the line segment PQ, then the locus of M intersects the latus rectum of the given ellipse at points

A
(±352,±27)
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B
(±352,±194)
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C
(±23,±17)
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D
(±23,±437)
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Solution

The correct option is C (±23,±17)
x2+4y2=16
x216+y24=1; a2=16, b2=4


P(4cosθ,2sinθ)
Equation of tangent at P:
xcosθ4+ysinθ2=1

Slope of tangent, mt=12cotθ
Slope of normal, mn=2tanθ

Equation of normal at P:
y2sinθ=2tanθ(x4cosθ)
Normal intersects the x-axis.
y=0
2sinθ=2sinθcosθ(x4cosθ) x=3cosθ
Therefore, coordinates of Q is (3cosθ,0)

P(4cosθ,2sinθ); Q(3cosθ,0)
M:(72cosθ,sinθ)
Let M be (h,k)
(27h)2+k2=1
x2(72)2+y21=1, which represents an ellipse.

Equation of latus rectum : x=±ae
e2=1b2a2=1416=1216
e=32
x=±23

Now,
x2(72)2+y21=1; x=±23
y=±17
Points of intersection (±23,±17)

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