Question

If the lines $x+y+1=0$ and $2x-y+2=0$ are the asymptotes of a hyperbola. If the line $x-2=0$ touches the hyperbola then the equation of the hyperbola is $4(x+y+1)(2x-y+2)=\lambda$. Find the value of $\lambda$.

Answer 1

Given asymptotes : $2x-y+2=0$ .......... $\left(1\right)$

$x+y+1=0$ ........... $\left(2\right)$

Tangent $T$ : $x-2=0$ ........ $\left(3\right)$

Equation of hyperbola: $4(x+y+1)(2x-y+2)=$$\lambda$ ......$\left(4\right)$

Comparing $T$ with $T:y=mx+c$

$\Rightarrow m=1,c=-2$

From equation $\left(3\right)$

$\Rightarrow$ Point of contact on tangent is $(2,0)$

Hence it lies on the hyperbola.

Putting $(2,0)$ in equation $\left(4\right)$

$\Rightarrow 4(2+0+1)\left(2\right(2)-0+2)=\lambda$

$\Rightarrow 72=\lambda$

$\Rightarrow \lambda =72$.