    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 4Let the point of concurrency be (h,k) The equation of normal at (x′,y′) is x−x′x′a2=y−y′y′b2 It passes through (h,k) Thus, (a2−b2)x′y′+b2kx′−a2hy′=0⋯(1) Equation (1) is satisfied by (x1,y1),(x2,y2),(x3,y3) and (x4,y4) Thus the curve passing through these points is (a2−b2)xy+b2kx−a2hy=0 From equation (1), y′=b2kx′a2h−(a2−b2)x′⋯(2) But, x′2a2+y′2b2=1⋯(3) Substituting equation (2) in equation (3), Thus eliminating y′ we get, −(a2−b2)x′4+2ha2(a2−b2)x′3+⋯−2a4h(a2−b2)x′+a6h2=0 Then x1+x2+x3+x4=2ha2a2−b2 x1x2x3+x2x3x4+x3x4x1+x1x2x4=−2a4h and, x1x2x3x4=a6h2−(a2−b2) So,(1x1+1x2+1x3+1x4)=2(a2−b2)a2h Hence, (x1+x2+x3+x4)(1x1+1x2+1x3+1x4)=4  Suggest Corrections  3      Similar questions  Related Videos   Ellipse and Terminologies
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