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Question

Let the normals at the four points (x1,y1),(x2,y2),(x3,y3) and (x4,y4) on the ellipse x2a2+y2b2=1 be concurrent at some point (called as conormal point). Then (x1+x2+x3+x4)(1x1+1x2+1x3+1x4) is equal to


A
4
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B
3
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C
4
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D
2
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Solution

The correct option is A 4
Let the point of concurrency be (h,k)
The equation of normal at (x,y) is xxxa2=yyyb2
It passes through (h,k)
Thus, (a2b2)xy+b2kxa2hy=0(1)
Equation (1) is satisfied by  (x1,y1),(x2,y2),(x3,y3) and (x4,y4)
Thus the curve passing through these points is (a2b2)xy+b2kxa2hy=0
From equation (1),
y=b2kxa2h(a2b2)x(2)
But, x2a2+y2b2=1(3)
Substituting equation (2) in equation (3), Thus eliminating y we get,
(a2b2)x4+2ha2(a2b2)x3+2a4h(a2b2)x+a6h2=0
Then x1+x2+x3+x4=2ha2a2b2
x1x2x3+x2x3x4+x3x4x1+x1x2x4=2a4h
and,  x1x2x3x4=a6h2(a2b2)
So,(1x1+1x2+1x3+1x4)=2(a2b2)a2h
Hence, (x1+x2+x3+x4)(1x1+1x2+1x3+1x4)=4

Mathematics

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