Let the point be $P(h, k)$.
Tangent from $(h, k)$ to hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\left(\frac{h x}{a^2}-\frac{k y}{b^2}-1\right)^2=\left(\frac{h^2}{a^2}-\frac{k^2}{b^2}-1\right)$
$$
\left(\frac{x^2}{a^2}-\frac{y^2}{b^2}-1\right)
$$


$$
\begin{array}{cc}
\Rightarrow & \frac{h^2 x^2}{a^4}+\frac{k^2 y^2}{b^4}+1-\frac{2 h k x y}{a^2 b^2}-\frac{2 h x}{a^2}+\frac{2 k y}{b^2} \\
= & \frac{h^2 x^2}{a^4}-\frac{h^2 y^2}{a^2 b^2}-\frac{h^2}{a^2}-\frac{x^2 k^2}{a^2 b^2}+\frac{y^2 k^2}{b^4}+\frac{k^2}{b^2}-\frac{x^2}{a^2}+\frac{y^2}{b^2}+1 \\
\Rightarrow \quad & \frac{x^2}{a^2}\left(\frac{k^2}{b^2}+1\right)-\frac{2 h k x y}{a^2 b^2}+\frac{y^2}{b^2} \\
& \left(\frac{h^2}{a^2}-1\right)-\frac{2 h x}{a^2}+\frac{2 k y}{b^2}+\frac{h^2}{a^2}=0 \\
\therefore & m_1 m_2=\frac{k^2+b^2}{h^2-a^2}
\end{array}
$$
Product of shape $=k^2$
$\therefore$ Locus of $p$ is $\frac{y^2+b^2}{x^2-a^2}=k^2$
$$
\Rightarrow \quad y^2+b^2=k^2\left(x^2-a^2\right)
$$