Example 2. The chord AB of the parabola $y=x^{2}$ keeps moving while being tangent to the circle $\odot O$ $x^{2}+y^{2}=1$. Find the locus of the intersection of the tangents to the parabola at points A and B.
Solution 1 As shown in Figure 3, let the chord AB of the parabola be tangent to $\odot \mathrm{O}$ at $\mathrm{P}(\alpha, \beta)$. Then the equation of the tangent line to $\odot \mathrm{O}$ at point $\mathrm{P}$ is
$$
\begin{array}{c}
\alpha \mathrm{x}+\beta \mathrm{y}=1, \\
\text { Let } \mathrm{A}\left(\mathrm{x}_{1}, \mathrm{x}_{1}^{2}\right), \\
\mathrm{B}\left(\mathrm{x}_{2}, \mathrm{x}_{2}^{2}\right), \\
\text { From }\left\{\begin{array}{l}
\alpha \mathrm{x}+\beta \mathrm{y}=1, \\
\mathrm{y}=\mathrm{x}^{2}
\end{array}\right. \text { we get } \\
\beta \mathrm{x}^{2}+\alpha \mathrm{x}-1=0,
\end{array}
$$

By Vieta's formulas, we have
$$
\mathrm{x}_{1}+\mathrm{x}_{2}=-\frac{\alpha}{\beta}, \mathrm{x}_{1} \cdot \mathrm{x}_{2}=-\frac{1}{\beta} .
$$

The equations of the tangents to the parabola $y=x^{2}$ at points $A$ and $B$ are $y+x_{1}^{2}=2 x_{1} x$ and $y+x_{2}^{2}=2 x_{2} x$, respectively. Therefore, the intersection point $Q(x, y)$ of the two tangents is
$$
\left\{\begin{array}{l}
y+x_{1}^{2}=2 x_{1} x, \\
y+x_{2}^{2}=2 x_{2} x
\end{array}\right.
$$
Subtracting (2) from (1) gives: $x_{1}^{2}-x_{2}^{2}=2\left(x_{1}-x_{2}\right) x$.

Thus, $2 \mathrm{x}=\mathrm{x}_{1}+\mathrm{x}_{2}=-\frac{\alpha}{\beta}$.
Multiplying (1) by $x_{2}$ and (2) by $x_{1}$ and then subtracting, we get: $\left(x_{2}-x_{1}\right) y+x_{1} x_{2}$ - $\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)=0$.
Thus, $\mathrm{y}=\mathrm{x}_{1} \mathrm{x}_{2}=-\frac{1}{\beta}$.
From (3) and (4), we get $\beta=-\frac{1}{\mathrm{y}}, \alpha=\frac{2 \mathrm{x}}{\mathrm{y}}$.
Since $\mathrm{P}(\alpha, \beta)$ lies on the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=1$,
$\therefore\left(\frac{2 x}{y}\right)^{2}+\left(\frac{1}{y}\right)^{2}=1$.
Simplifying, we get: $y^{2}-4 x^{2}=1$.
By the given conditions, the required trajectory is part of the lower branch of a hyperbola with foci on the $\mathrm{y}$-axis, $a=1$, and $b=\frac{1}{2}$.

Solution 2 As shown in Figure 3, let the chord $\mathrm{AB}$ of the parabola be tangent to $\odot$ O at $\mathrm{P}(\alpha, \beta)$. Then the equation of the tangent line $\mathrm{AB}$ to $\odot \mathrm{O}$ at point $\mathrm{P}$ is: $\alpha \mathrm{x}+\beta \mathrm{y}-1=0$.

Let the intersection point of the tangents to the parabola at points $A$ and $B$ be $Q\left(x_{0}, y_{0}\right)$. Then the equation of the chord of contact $AB$ is:
$y+y_{0}=2 x_{0} x$, i.e., $2 x_{0} x-y-y_{0}=0$. (2)
Since (1) and (2) are the same line, by the necessary and sufficient condition for two lines to coincide, we have:
$$
\frac{\alpha}{2 x_{0}}=\frac{\beta}{-1}=\frac{1}{y_{0}} \text {. }
$$

Thus, $\alpha=\frac{2 x_{0}}{\mathrm{y}_{0}}, \quad \beta=\frac{-1}{\mathrm{y}_{0}}$.
Since $\mathrm{P}(\alpha, \beta)$ lies on the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=1$,
$\therefore\left(\frac{2 x_{0}}{y_{0}}\right)^{2}+\left(-\frac{1}{y_{0}}\right)^{2}=1$.
Simplifying, we get: $y_{0}^{2}-4 x_{0}^{2}=1$,
i.e., $y^{2}-4 x^{2}=1$.