Question

Consider a hyperbola whose centre is at origin. If 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, then the equation of the tangent to the hyperbola at $(-1,\frac{7}{2})$ is

• A. $x-2y=-8$
• B. $2x-y=-\frac{11}{2}$
• C. $x+2y=6$
• D. $3x+2y=4$

Answer 1

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

We know that in a hyperbola, the portion of tangent intercepted between asymptotes is bisected at the point of contact.
The equation of tangent in parametric form is given by
$\frac{x-1}{-\frac{1}{\sqrt{2}}}=\frac{y-1}{\frac{1}{\sqrt{2}}}=\pm 3\sqrt{2}$
$\Rightarrow A\equiv (4,-2),B\equiv (-2,4)$

The equations of asymptotes $(OA$ and $OB)$ are given by
$y+2=\frac{-2}{4}(x-4)$
$\Rightarrow 2y+x=0$
and $y-4=\frac{4}{-2}(x+2)$
$\Rightarrow 2x+y=0$
Hence, the combined equation of asymptotes is
$(2x+y)(x+2y)=0$
$\Rightarrow 2x^2+2y^2+5xy=0$

Let the equation of the hyperbola be
$2x^2+2y^2+5xy+\lambda =0$
It passes through $(1,1)$. Therefore,
$2+2+5+\lambda =0$
$\Rightarrow \lambda =-9$
So, the hyperbola is
$2x^2+2y^2+5xy=9$
The equation of the tangent at $(-1,\frac{7}{2})$ is given by
$2x(-1)+2y\left(\frac{7}{2}\right)+5\times \frac{x\left(\frac{7}{2}\right)+(-1)y}{2}=9$
$\Rightarrow 3x+2y=4$