Let a tangent to the curve   y 2   =   24 x meet the curve x y   =   2 at the…

Given:

$y^2=24x$

Comparing with $y^2=4ax$, we get

$4a=24\Rightarrow a=6$

Also given $xy=2$

Let any point $Q\equiv \left(a{t}^2,2at\right)\equiv \left(6t^2,12t\right)$ on the parabola.

Equation of tangent at $Q$ is

$12yt=12\left(x+6t^2\right)$

$\Rightarrow yt=x+6t^2$

$\Rightarrow x-yt+6t^2=0 ....\left(\mathrm{i}\right)$

Let mid-point of $AB$ be $P\left(h,k\right)$.

Equation of chord of hyperbola is

$\frac{xk+yh}{2}=hk$

$\Rightarrow xk+hy-2hk=0 ...\left(\mathrm{ii}\right)$

Since, $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right)$ are same, so on comparing, we get

$\frac{k}{1}=\frac{h}{-t}=\frac{-2hk}{6t^2}$

$\Rightarrow t=-\frac{h}{k} \&$$\frac{h}{-t}=\frac{-2hk}{6t^2}\Rightarrow t=\frac{k}{3}$

So,

$\frac{k}{3}=-\frac{h}{k}$

^{$\Rightarrow k^2=-3h$}

Hence, locus is $y^2=-3x$.

Therefore, directrix is $4x=3$.

Length of latus rectum is $=4a=3$