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Question

The slope(s) of the common tangent(s) to the two hyperbolas x2a2y2b2=1 and y2a2x2b2=1 is/are

A
1
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B
2
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C
1
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D
12
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Solution

The correct option is C 1
Let the slope of tangent is m, then
Tangent to x2a2y2b2=1 is
y=mx±a2m2b2(1)
The tangent to other hyperbola
y2a2x2b2=1 is
y=mx±a2b2m2(2)
both are same equations
a2m2b2=a2b2m2
(a2+b2)m2=a2+b2
m=±1

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