## Task B-4.2.

The lines with equations $y=k_{1} x+2$ and $y=k_{2} x-3$ are tangents to the parabola $\mathrm{s}$ with the equation $y^{2}=2 p x, p>0$, and they enclose an angle of $45^{\circ}$. Determine the equations of all such parabolas.
## First Solution.

Given that the angle between two lines is $45^{\circ}$, it follows that $\left|\frac{k_{2}-k_{1}}{1+k_{1} k_{2}}\right|=1$ or $\left|k_{2}-k_{1}\right|=$ $\left|1+k_{1} k_{2}\right|$.

From the condition of tangency $p=2 k l$, we get $4 k_{1}=-6 k_{2}$, i.e., $k_{1}=-\frac{3}{2} k_{2}$.

Thus, $\left|k_{2}+\frac{3}{2} k_{2}\right|=\left|1-\frac{3}{2} k_{2} \cdot k_{2}\right|$, i.e., $\left|\frac{5}{2} k_{2}\right|=\left|1-\frac{3}{2} k_{2}^{2}\right|$.

We have two cases:

$$
\frac{5}{2} k_{2}=1-\frac{3}{2} k_{2}^{2} \quad \text{and} \quad -\frac{5}{2} k_{2}=1-\frac{3}{2} k_{2}^{2}
$$

Therefore, there are 4 solutions.

$$
\begin{array}{llll}
k_{1}=-\frac{1}{2} & k_{1}=3 & k_{1}=\frac{1}{2} & k_{1}=-3 \\
k_{2}=\frac{1}{3} & k_{2}=-2 & k_{2}=-\frac{1}{3} & k_{2}=2
\end{array}
$$

The corresponding parameters are

$$
p=-2 \quad p=12 \quad p=2 \quad p=-12
$$

Due to the condition $p>0$, the solutions to the problem are $y^{2}=4 x$ and $y^{2}=24 x$.

Note: It is not necessary for the student to determine both coefficients $k_{1}$ and $k_{2}$. It is sufficient to calculate the parameter $p$ using one of these coefficients.