## Task A-4.6.

Given is the point $A(0,2)$ on the parabola $y^{2}=x+4$. Find all points $B$ different from $A$ on the given parabola for which there exists a point $C$, also on the parabola, such that the angle $\varangle A C B$ is a right angle.
## Solution.

Let $B\left(b^{2}-4, b\right)$ and $C\left(c^{2}-4, c\right)$ for some real numbers $b$ and $c$. Note that it is impossible for $c = \pm 2$ or $c = \pm b$ because otherwise, some two of the points $A, B$, and $C$ would coincide, or they would lie on a line parallel to the $x$-axis.
![](https://cdn.mathpix.com/cropped/2024_05_30_4cdaff2107ef283b1d94g-24.jpg?height=408&width=1522&top_left_y=700&top_left_x=196)

The slopes of the lines $A C$ and $B C$ are, respectively,

$$
\frac{c-2}{c^{2}-4}=\frac{1}{c+2} \quad \text { and } \quad \frac{c-b}{c^{2}-b^{2}}=\frac{1}{c+b}
$$

The lines $A C$ and $B C$ are perpendicular, which means that

$$
\frac{1}{c+2} \cdot \frac{1}{c+b}=-1
$$

or, equivalently,

$$
c^{2}+b c+2 c+2 b+1=0
$$

We want to find all points $B$ on the parabola for which there exists a point $C$ on the parabola such that the angle $\varangle A C B$ is a right angle, which is equivalent to finding real numbers $b$ for which the quadratic equation in $c$

$$
c^{2}+(b+2) c+2 b+1=0
$$

has real solutions. For this, it is necessary and sufficient that the discriminant of this equation is non-negative, i.e.,

$$
(b+2)^{2}-4(2 b+1) \geqslant 0
$$

This is satisfied by numbers $b$ for which $b(b-4) \geqslant 0$, i.e., $b \leqslant 0$ or $b \geqslant 4$.

The desired points are all points of the form $B=\left(b^{2}-4, b\right)$, where $b \in\langle-\infty, 0] \cup[4,+\infty\rangle$.

![](https://cdn.mathpix.com/cropped/2024_05_30_4cdaff2107ef283b1d94g-24.jpg?height=391&width=905&top_left_y=2192&top_left_x=513)