Example 2. The tangent line at point $\mathrm{P}_{0}\left(x_{0}, y_{0}\right)$ on the ellipse $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$ intersects the hyperbola $b^{2} x^{2}-a^{2} y^{2}=a^{2} b^{2}$ at points $P_{1}$ and $P_{2}$. Find the equation of the locus of the intersection point $\mathrm{P}^{\prime}$ of the tangents to the hyperbola at $P_{1}$ and $P_{2}$.
From the problem, we know that the tangent line of the ellipse at $\mathrm{P}_{0}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$ is exactly the chord of the tangent points of the hyperbola about the moving point $\mathrm{P}^{\prime}$. Thus, the equation 
$$
b^{2} x_{0} x+a^{2} y_{0} y
=a^{2} b^{2}
$$ 
should be consistent with 
$$
b^{2} x^{\prime} x
-a^{2} y^{\prime} y=a^{2} b^{2}.
$$ 
Therefore, we have
$$
\frac{\mathrm{b}^{2} \mathrm{x}_{0}}{\mathrm{~b}^{2} \mathrm{x}^{\prime}}=\frac{\mathrm{a}^{2} \mathrm{y}_{0}}{-\mathrm{a}^{2} \mathrm{y}^{\prime}}=\frac{\mathrm{a}^{2} \mathrm{~b}^{2}}{\mathrm{a}^{2} \mathrm{~b}^{2}},
$$

which implies $\frac{x_{0}}{x^{\prime}}=\frac{-y_{0}}{y^{\prime}}=1$.

On the ellipse, we can deduce that $b^{2} x^{\prime 2}+a^{2} y^{\prime 2}=a^{2} b^{2}$. Therefore, the trajectory equation of $\mathrm{P}^{\prime}$ is the ellipse equation $b^{2} x^{2}+a^{2} y^{2}$ $=\mathrm{a}^{2} \mathrm{~b}^{2}$. (See Figure 1)