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Question

The line lx+my=1 meets the ellipse x2a2+y2b2=1 in the points P and Q, then the mid point of chord PQ is, where k=a2l2+b2m2

A
(a2lk,b2mk)
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B
(a2lk,b2mk)
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C
(b2lk,a2mk)
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D
(b2lk,a2mk)
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Solution

The correct option is D (a2lk,b2mk)
x2a2+y2b2=1
Or
b2x2+a2y2=a2b2
Now the line lx+my1 cuts the ellipse at two points.
y=1lxm
Substituting in the equation of ellipse gives us
b2x2+a2(1lxm)2=a2b2
m2b2x2+a2(1lx)2=a2b2m2
m2b2x2+a2[1+l2x22lx]=a2b2m2
x2(m2b2+a2l2)2a2lx+a2a2b2m2=0
x=2a2l±4a4l24(m2b2+a2l2)(a2a2b2m2)2(m2b2+a2l2)
x=a2l±a4l2(m2b2+a2l2)(a2a2b2m2)(m2b2+a2l2)
x=a2l±a4l2a2(a2l2+m2b2)(1m2b2)(m2b2+a2l2)
Let the xcoordinate of P and Q be x1,x2, then
x1+x2=BA=2a2la2l2+m2b2
Hence
x1+x22=a2la2l2+m2b2
x=a2la2l2+m2b2
Thus y=1lxm
=1l(a2la2l2+m2b2)m
=a2l2+m2b2a2l2m(a2l2+m2b2)
=m2b2m(a2l2+m2b2)
=b2ma2l2+m2b2
Hence
(x,y)=(a2la2l2+m2b2,b2ma2l2+m2b2)

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