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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 C

(±23,±17)


Given, x216+y24=1
Here, a = 4, b = 2
Equation of normal
4xsecθ2ycosecθ=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=1416=34
e=32
x=±4×32=±23[x=±ae] ... (iv)
On solving Eqs. (iii) and (iv), we get
449×12+y2=1y2=14849=149
y=±17
Required points (±23,±17).


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