The locus of the point of intersection of perpendicular tangents to the ellipse is called

The equation of a pair of tangents from $(\alpha, \beta)$ to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $$ \left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-1\right)\left(\frac{\alpha^{2}}{a^{2}}+\frac{\beta^{2}}{b^{2}}-1\right)=\left(\frac{x \alpha}{a^{2}}+\frac{y \beta}{b^{2}}-1\right)^{2} $$ $\left(\mathrm{SS}_{1}=\mathrm{T}^{2}\right)$ The tangents are perpendicular, if the coefficient of $x^{2}+$ the coefficient of $y^{2}=0$. $$ \begin{array}{r} \Rightarrow \frac{1}{a^{2}}\left(\frac{\alpha^{2}}{a^{2}}+\frac{\beta^{2}}{b^{2}}-1\right)-\frac{\alpha^{2}}{a^{4}}+\frac{1}{b^{2}}\left(\frac{\alpha^{2}}{a^{2}}+\frac{\beta^{2}}{b^{2}}-1\right) \\ -\frac{\beta^{2}}{b^{4}}=0 \end{array} $$ $$ \begin{gathered} \Rightarrow \quad \frac{\beta^{2}}{a^{2} b^{2}}-\frac{1}{a^{2}}+\frac{\alpha^{2}}{a^{2} b^{2}}-\frac{1}{b^{2}}=0 \\ \Rightarrow \quad \alpha^{2}+\beta^{2}=a^{2} b^{2}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right) \\ =a^{2} b^{2}\left(\frac{a^{2}+b^{2}}{a^{2} b^{2}}\right) \\ =a^{2}+b^{2} \end{gathered} $$ Hence, the locus of $(\alpha, \beta)$ is the circle $$ x^{2}+y^{2}=a^{2}+b^{2} $$ This circle is called the director circle.