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 $AB=6\sqrt{2}$ units.

Equation of the tangent to the hyperbola at $(-1,\frac{7}{2})$ is

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

Answer 1

The correct option is B $3x+2y=4$

Let the equaiton 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 $(-1,\frac{7}{2})$ is given by
$2x(-1)+2y\left(\frac{7}{2}\right)+5\frac{x\left(\frac{7}{2}\right)+(-1)y}{2}=9$
or $3x+2y=4$