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Question

A normal to x2a2+y2b2=1 meets the axes in L and M. The perpendiculars to the axes through L and M intersect at P. Then the equation to the locus of P is


Your Answer
A
a2x2b2y2=(a2+b2)2
Your Answer
B
a2x2+b2y2=(a2+b2)2
Your Answer
C
b2x2a2y2=(a2b2)2
Correct Answer
D
a2x2+b2y2=(a2+b2)2

Solution

The correct option is D a2x2+b2y2=(a2+b2)2
P=(x1,y1),xx1+yy1=1 Apply normal condition

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