    Question

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

A
(±357,±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 (±2√3,±17)Given Ellipse x216+y24=1e=√1−b2a2=√32∵P is a point on the ellipse So, P=(4cosθ,2sinθ)Equation of normal to the ellipse x216+y24=1 at point (x1,y1)=(4cosθ,2sinθ) is given bya2y1(x−x1)=b2x1(y−y1) ⟹16×2sinθ(x−4cosθ)=4×4cosθ(y−2sinθ)⟹2xsinθ−8sinθcosθ=ycosθ−2sinθcosθ⟹2xsinθ=ycosθ+6sinθcosθ⟹2xcosθ=ysinθ+6⟹2xsecθ−ycosecθ=6It meet the x-axis at Q(3cosθ,0) ∴M=(72cosθ,sinθ)=(x,y)Locus of M is x2(72)2+y21=1Latus rectum of the given ellipse isx=±ae=±√16−4=±2√3So locus of M meets the latus rectum at points for whichy2=1−12×449=149 ⇒ y=±17Hence, the required point is (±2√3,±17).  Suggest Corrections  0      Similar questions
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