In a hyperbola, the portion of the tangent intercepted between the asymptotes is bisected at the point of contact.
Consider a hyperbola whose center is at the origin. A line `x+y=2` touches this hyperbola at P(1,1) and intersects the asymptotes at A and B such that AB = `6sqrt2` units.
The equation of the tangent to the hyperbola at `(-1, 7//2)` is

Options

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

Answer

B

Explanation

Let the equation of the hyperbola be
`2x^(2)+2y^(2)+5xy+lambda=0`
It passes through (1,1). Therefore,
`2+2+5+lambda=0`
`"or "lambda=-9`
So, the hyperbola is
`2x^(2)+2y^(2)+5xy=9`
The equation of the tangent at `(-1,7//2)` is given by
`2x(-1)+2y((7)/(2))+5(x(7//2)+(-1)y)/(2)=9`
`"or "3x+2y=4`