From a point on the line x - y + 2 = 0 tangents are drawn to the...

Let the point be $\left(\alpha, \beta\right) ⇒ \beta = \alpha + 2$
Chord of contact of the hyperbola is $T = 0$
$⇒ \dfrac{\alphax}{6} - \dfrac{\betay}{2} = 1$
$⇒ \dfrac{\alphax}{6} - \dfrac{\left(\alpha + 2\right)y}{2} = 1$
$⇒ \alpha\left(\dfrac{x}{6} - \dfrac{y}{2}\right) - \left(\dfrac{2y}{2} +1\right) = 0$
Since, this passes through the fixed point \left(\lambda, \mu\right)$<br />$∴ \dfrac{\lambda}{6} - \dfrac{\mu}{2} =0 and $\dfrac{2\mu}{2} +1 = 0$

So, $\mu = - 1$ and $\lambda = - 3$
$⇒ \mu - \lambda = 2$