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. `5xy+2x^(2)+2y^(2)=0`
B. `3x+2y=4`
C. `2x^(2)+2y^(2)-5xy=0`
D. none of these

Answer

A

Explanation

The equation of tangent in parametric form is given by
`(x-1)/(-1//sqrt2)=(y-1)/(1//sqrt2)= pm3sqrt2`
`"or "A-=(4,-2), B-=(-2,4)`
The equation of asymptotes (OA and OB) are given by
`y+2=(-2)/(4)(x-4)`
`"or "2y+x=0`
`"and "y-4=(4)/(-2)(x+2)`
`"or "2x+y=0`
Hence, the combined equation of asymptotes is
`(2x+y)(x+2y)=0`
`"or "2x^(2)+2y^(2)+5xy=0`