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Question

In a hyperbola portion of tangent intercepted between asymptotes is bisected at the point of contact. Consider a hyperbola whose centre is at origin. A line x+y=2 touches this hyperbola at P(1,1) and intersects the asymtotes at A and B such that AB=62 units.

Equation of the tangent to the hyperbola at (1,72) is

A
5x+2y=2
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B
3x+2y=4
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C
3x+4y=11
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D
none of these
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Solution

The correct option is B 3x+2y=4
Let the equaiton of the hyperbola be
2x2+2y2+5xy+λ=0
It passes through (1, 1). Therefore,
2+2+5+λ=0
or λ=9
So, the hyperbola is
2x2+2y2+5xy=9
The equation of the tangent at (1,72) is given by
2x(1)+2y(72)+5x(72)+(1)y2=9
or 3x+2y=4

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