1. Find the locus of the intersection point of two perpendicular intersecting tangents to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
This problem can be solved in several ways. When explaining this problem, I combine various methods provided by students to give the following solution:
Given the equation of the tangent line with a known slope:
$$
\begin{aligned}
& y=k x \pm \sqrt{a^{2} k^{2}-b^{2}} \\
\Longrightarrow & y-k x= \pm \sqrt{a^{2} k^{2}}-b^{2} \\
\Rightarrow & \left(a^{2}-x^{2}\right) k^{2}+2 x k+b^{2}-y^{2}=0 .
\end{aligned}
$$

Since the tangents are perpendicular to each other, and according to Vieta's formulas, we have
$$
\begin{aligned}
\frac{b^{2}-y^{2}}{a^{2}-x^{2}}=-1 \\
\Rightarrow x^{2}+y^{2}=a^{2}+b^{2} .
\end{aligned}
$$

This is the locus of the intersection points of the two tangents. Some students are half-convinced and half-doubtful, finding it hard to understand. The fundamental reason is the lack of deep understanding of the equation and the meaning of the parameter \( k \). In fact, the equation \(\left(a^{2}-x^{2}\right) k^{2}+2 x k+b^{2}-y^{2}=0\) is the condition that any tangent slope \( k \) must satisfy, and there are only two roots for \( k \) in the equation. These two roots are the slopes of the two tangents drawn through the point \((x, y)\). Since the two tangents are perpendicular, according to Vieta's formulas, the equation of the circle can be derived. If the idea of combining numbers and shapes is not used to understand \( k \), it would naturally be difficult to grasp the essence of this solution.