1. Determine the geometric locus of points $M$ from which the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is viewed at a right angle.
Solution. Let $M(\alpha, \beta)$ be a point from which the given ellipse is viewed at a right angle. An arbitrary line through $M$ has the equation

$$
y-\beta=k(x-\alpha), \text { i.e. } y=k x+\beta-k \alpha
$$

or $x=\alpha$. The condition for the line $y=k x+\beta-k \alpha$ and the ellipse to have exactly one common point is that the system of equations in $x$ and $y$

$$
y=k x+\beta-k \alpha, \quad b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}
$$

has a unique solution, i.e., the equation

$$
\left(b^{2}+a^{2} k^{2}\right) x^{2}+2 a^{2} k(\beta-k \alpha) x+a^{2}(\beta-k \alpha)^{2}-a^{2} b^{2}=0
$$

has a unique solution. Therefore, it is necessary and sufficient for the discriminant to be zero, i.e.,

$$
\left(a^{2}-\alpha^{2}\right) k^{2}-2 \beta \alpha k+b^{2}-\beta^{2}=0
$$

Let $k_{1}$ and $k_{2}$ be the solutions of this quadratic equation. For the ellipse to be viewed at a right angle from $M$, it is necessary for the lines $y=k_{1} x+\beta-k_{1} \alpha$ and $y=k_{2} x+\beta-k_{2} \alpha$ to intersect at a right angle, i.e., $k_{1} k_{2}=-1$. However, $k_{1} k_{2}=\frac{b^{2}-\beta^{2}}{a^{2}-\alpha^{2}}$, from which it follows that $b^{2}-\beta^{2}=-a^{2}+\alpha^{2}$, i.e., $\alpha^{2}+\beta^{2}=a^{2}+b^{2}$. If the line $x=\alpha$ is tangent to the ellipse, then $\alpha= \pm a$ and for the ellipse to be viewed at a right angle from $M$, it must be $\beta= \pm b$. Thus, in this case as well, $\alpha^{2}+\beta^{2}=a^{2}+b^{2}$. Therefore, the required geometric locus is a circle with center at $(0,0)$ and radius $\sqrt{a^{2}+b^{2}}$.