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Question

Show that the curves x2a2+λ1+y2b2+λ1=1 and x2a2+λ2+y2b2+λ2=1intersect at right angles.

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Solution

We have, x2a2+λ1+y2b2+λ1=1 ...1and x2a2+λ2+y2b2+λ2=1 ...2Now we can find the slope of both the curve by differentiating w.r.t x2xa2+λ1+2ydydxb2+λ1=0 and 2xa2+λ2+2ydydxb2+λ2=0dydx=-xy×b2+λ1a2+λ1 and dydx=-xy×b2+λ2a2+λ2m1=-xy×b2+λ1a2+λ1 and m2=-xy×b2+λ2a2+λ2Subtracting 2 from 1, we get,x21a2+λ1-1a2+λ2+y21b2+λ1-1b2+λ2=0x2y2=λ2-λ1b2+λ1b2+λ2×1λ1-λ2a2+λ1a2+λ2Now,m1××m2=x2y2×b2+λ1a2+λ1×b2+λ2a2+λ2 =λ2-λ1b2+λ1b2+λ2×a2+λ1a2+λ2λ1-λ2×b2+λ1a2+λ1×b2+λ2a2+λ2 =-1hence, 1 and 2 cuts orthogonally.

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