Parabola’s $y^2=4a(x-k)$ and $x^2=4a(y-l)$ are such that they touch each other (k,l variables). Locus of point of contact is $xy=m{a}^2$ Then m=

The given curves are $y^2=4a(x-k)$ and $x^2=4a(y-l)$

Syppose that the curves are touch each other at a point $P\left(x_1,y_1\right)$

Hence, the slopes of tangents at $P\left(x_1,y_1\right)$ to the curves are equal. 

Hence, 

$2y\frac{dy}{dx}=4a\Rightarrow \frac{dy}{dx}=\frac{2a}{y}$

So that $m_1=\frac{2a}{y_1}$

and $m_2=\frac{x_1}{2a}$

Since the curves touch each other, their slopes must be equal at their point of contact

hence, 

$\frac{2a}{y_1}=\frac{x_1}{2a}$

Cross multiply

$x_1{y}_1=4a^2$

Threfore, the locus of point of contact is $xy=4a^2$

Therefore, $m=\boxed{4}$