    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 the locus of M intersects the latusrectum of the given ellipse at the 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 D (±2√3,±17) Given, x216+y24=1 Here, a = 4, b = 2 Equation of normal 4xsecθ−2ycosecθ=12 M(7cosθ2,sinθ)=(h,k) [say] ∴h=7cosθ2⇒=2h7=cosθ ... (i) and k=sinθ ...(ii) On squaring and adding Eqs. (i) and (ii), we get 4h249+k2=1 [∵cos2θ+sin2θ=1] Hence, locus is 4x249+y2=1 ...(iii) For given ellipse, e2=1−416=34 ∴e=√32 ∴x=±4×√32=±2√3[∵x=±ae] ... (iv) On solving Eqs. (iii) and (iv), we get 449×12+y2=1⇒y2=1−4849=149 y=±17 ∴ Required points (±2√3,±17).  Suggest Corrections  0      Similar questions  Related Videos   Line and Ellipse
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