Given that, parabola $\mathrm{x}^{2}=-8 \mathrm{y}$



Here, on comparing with $\mathrm{x}^{2}=4 \mathrm{ay}$
$$
\Rightarrow \quad 4 \mathrm{a}=-8 \Rightarrow \mathrm{a}=-2
$$
Let the parametric coordinate of parabola $\mathrm{x}^{2}=-8 \mathrm{y}$ is, $\mathrm{P} \rightarrow\left(4 \mathrm{t},-2 \mathrm{t}^{2}\right)$
and the other coordinate of latusrectum is $\mathrm{P}^{\prime} \rightarrow\left(\frac{-4}{\mathrm{t}}, \frac{-2}{\mathrm{t}^{2}}\right)$
Now, the equation tangent of parabola $x^{2}=-8 y$
At P $\quad \begin{aligned} \mathrm{x} \cdot \mathrm{x}_{1} &=-4\left(\mathrm{y}+\mathrm{y}_{1}\right) \quad \text{...(i)} \\ \mathrm{x}(4 \mathrm{t}) &=-4\left(\mathrm{y}-2 \mathrm{t}^{2}\right) \\ \mathrm{x} t &=-\mathrm{y}+2 \mathrm{t}^{2} \\ \mathrm{xt} &+\mathrm{y}=2 \mathrm{t}^{2} \quad \text{...(ii)} \end{aligned}$
At $\mathrm{P}^{\prime} \quad \mathrm{x}\left(\frac{-4}{\mathrm{t}}\right)=-4\left(\mathrm{y}-\frac{2}{\mathrm{t}^{2}}\right)$
$\Rightarrow \quad \frac{x}{t}=y-\frac{2}{t^{2}}$
$\Rightarrow \quad x t=\mathrm{yt}^{2}-2$
$\Rightarrow \quad \mathrm{xt}-\mathrm{yt}^{2}=-2 \quad \text{...(iii)}$
On solving Eqs. (i) and (ii)
$$
\begin{aligned}
\mathrm{xt}^{3}+\mathrm{yt}^{2} &=2 \mathrm{t}^{4} \\
\mathrm{xt}-\mathrm{yt}^{2} &=-2 \\
\mathrm{xt}\left(1+\mathrm{t}^{2}\right) &=-2\left(1-\mathrm{t}^{4}\right) \\
\mathrm{xt}\left(1+\mathrm{t}^{2}\right) &=-2\left(1+\mathrm{t}^{2}\right)\left(1-\mathrm{t}^{2}\right) \\
\mathrm{tx} &=-2\left(1-\mathrm{t}^{2}\right) \quad \text{...(iv)}
\end{aligned}
$$
From Eq. (ii)
$$
\begin{aligned}
&-2\left(1-t^{2}\right)+y &=2 t^{2} \\
\Rightarrow & &-2+2 t^{2}+y &=2 t^{2} \\
\Rightarrow & & y &=2
\end{aligned}
$$
Hence, the intersection point of both tangent lying on Q.
ie, $\mathrm{y}=2$. Which is the required locus.