The equation of any tangent to $y^2=4ax$
$y= mx +\large\frac{a}{m}$
If it touches $x^2=4ay$ then, the equation.
$x62=4a \bigg( mx +\large\frac{a}{m}\bigg)$ has equal roots.
=> $mx^2-4am^2x-4a^2=0$
it has equal roots
=> $16a^2m^4=-16 a^2m$
=> $m=-1$
Putting $m=-1$ in $y= mx +\large\frac{a}{m}$ we get,
$y= -x-a$ or $x+y+a=0$
Hence A is the correct answer.