## Task B-4.4.

Determine the set of all points from which the hyperbola $16 x^{2}-25 y^{2}=400$ is seen at a right angle.
## Solution.

We are looking for the set of points where the tangents to the hyperbola are perpendicular.

Let the tangent line $\mathrm{s}$ have the equation $y=k x+l$.

Using the condition of tangency of the line and the hyperbola, and $l=y-k x$, we get

$$
25 k^{2}-16=(y-k x)^{2}
$$

After squaring, we rewrite the obtained equation $25 k^{2}-16=y^{2}-2 k x y+k^{2} x^{2}$ as a quadratic equation in $k$:

$$
\left(25-x^{2}\right) k^{2}+2 x y k-16-y^{2}=0 \text {. }
$$

Since the tangents must be perpendicular, we have $k_{1} \cdot k_{2}=-1$.

From Vieta's formulas, it follows that

$$
\frac{-16-y^{2}}{25-x^{2}}=-1 \Rightarrow x^{2}+y^{2}=9
$$

All the desired points lie on a circle centered at the origin with a radius of 3.