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Q.

Equation of a common tangent to the two hyperbolas x2a2y2b2=1and y2a2x2b2=1is

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a

y=xa2b2

b

y=xa2b2

c

y=x+a2b2

d

y=x+a2b2

answer is A, B, C, D.

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Detailed Solution

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x2a2y2b2=1     Andy2a2x2b2=1

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Tangent  toy=mx±a2m2b2

If this is also tangent to x2(b)2y2(a)2=1

a2m2+b2=(b2)m2(a2)=a2b2m2

(a2b2)m2=a2b2m=1

Hence four common tangent are y=±x±a2b2

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