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Question

# Two curves aix2+biy2=1; i=1,2 where a1≠a2, b1≠b2, a1,a2,b1,b2≠0 may intersect orthogonally if _____

A
a1a2=b1b2
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B
a11a12=b11b12
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C
a11+a12=b11+b12
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D
a1b2=a2b1
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Solution

## The correct option is B a−11−a−12=b−11−b−12The curves are a1x2+b1y2=1 ⋯(1) and a2x2+b2y2=1 ⋯(2) Differentiating both equations (dydx)1=−a1xb1y (dydx)2=−a2xb2y The curves intersect orthogonally. ⇒(−a1xb1y)(−a2xb2y)=−1 ⇒a1a2x2=−b1b2y2 ⋯(3) (1)−(2) gives (a1−a2)x2+(b1−b2)y2=0 ⇒(a1−a2)−b1b2y2a1a2+(b1−b2)y2=0 (From (3)) ⇒a1−a2a1a2=b1−b2b1b2 ⇒1a2−1a1=1b2−1b1 ⇒1a1−1a2=1b1−1b2

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