Given that, parabola $x^{2}=-8 y$
Here, on comparing with $x^{2}=4 a y$
$\Rightarrow 4 a=-8 \Rightarrow a=-2$
Let the parametric coordinate of parabola
$x^{2}=-8 y$ is, $P \rightarrow\left(4 t,-2 t^{2}\right)$
and the other coordinate of latusrectum is
$P' \rightarrow\left(\frac{-4}{t}, \frac{-2}{t^{2}}\right)$
Now, the equation tangent of parabola $x^{2}=-8 y$
$x \cdot x_{1}=-4\left(y+y_{1}\right)$...(i)
At $P x(4 t)=-4\left(y-2 t^{2}\right)$
$x t=-y+2 t^{2}$
$x t+ y=2 t^{2}$...(ii)
$\Rightarrow x\left(\frac{-4}{t}\right)=-4\left(y-\frac{2}{t^{2}}\right)$
$\Rightarrow \frac{x}{t}=y-\frac{2}{t^{2}}$
$\Rightarrow x t=y t^{2}-2$
$\Rightarrow x t-y t^{2}=-2$...(iii)
On solving Eqs. (i) and (ii)
$x t^{3}+y t^{2} =2 t^{4}$
$x t-y t^{2} =-2$
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$x t\left(1+t^{2}\right) =-2\left(1-t^{4}\right)$
$x t\left(1+t^{2}\right) =-2\left(1+t^{2}\right)\left(1-t^{2}\right)$
$t x =-2\left(1-t^{2}\right)$...(iv)
From Eq. (ii)
$-2\left(1-t^{2}\right)+y =2 t^{2}$
$\Rightarrow -2+2 t^{2}+y =2 t^{2}$
$\Rightarrow y =2$
Hence, the intersection point of both tangent lying on $Q$.
ie, $y=2 .$ Which is the required locus.
