Let's find the tangents to the parabola given by the equation $y=x^{2}$ that form a $45^{\circ}$ angle with the line segment drawn from the focus to the point of tangency.
We know that the curve of the quadratic function $y=\frac{1}{2 p} x^{2}$ is a parabola whose axis of symmetry is the $y$-axis, its vertex is at the origin, and it opens upwards. The parameter of the parabola is $p$, which measures the distance from the focus to the directrix. In our case, $y=x^{2}$, i.e., $\frac{1}{2 p}=1, p=\frac{1}{2}$, and the coordinates of the focus are $F\left(0 ; \frac{1}{4}\right)$. Let $E$ be a point of tangency (we will only consider tangents in the first quadrant for now), with coordinates $\left(x_{0} ; x_{0}^{2}\right)$. We seek the equation of the tangent $e$ that forms a $45^{\circ}$ angle with the line $E F$.

First, let's write the equation of the line $E F$. A direction vector of $E F$ has coordinates $V \left(x_{0} ; x_{0}^{2}-\frac{1}{4}\right)$; a point on the line has coordinates $F\left(0 ; \frac{1}{4}\right)$, so the equation of $E F$ is:

$$
y=\frac{x_{0}^{2}-\frac{1}{4}}{x_{0}} x+\frac{1}{4}
$$

It is known that the general equation of the tangent to a parabola is:

$$
x x_{0}=p\left(y+y_{0}\right)
$$

where $x_{0}, y_{0}$ are the coordinates of the point of tangency and $p$ is the parameter.

In our case, $p=\frac{1}{2}$, so the equation of the tangent is: $x x_{0}=\frac{1}{2}\left(y+x_{0}^{2}\right)$. Rearranging:

$$
y=2 x_{0} x-x_{0}^{2}
$$

If the lines with slopes $m_{1}, m_{2}$ are not perpendicular to each other, then the angle $\alpha$ between them satisfies the equation:

$$
\operatorname{tg} \alpha=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|
$$

(See, for example, page 339 of György Hajós's "Introduction to Geometry".)

$$
\operatorname{tg} 45^{\circ}=1, \quad m_{1}=\frac{x_{0}^{2}-\frac{1}{4}}{x_{0}}, \quad m_{2}=2 x_{0}
$$

Substituting these values, we get:

$$
1=\left|\frac{\frac{x_{0}^{2}-\frac{1}{4}}{x_{0}}-2 x_{0}}{1+\frac{x_{0}^{2}-\frac{1}{4}}{x_{0}} \cdot 2 x_{0}}\right|
$$

Considering $x_{0}>0$ and simplifying:

$$
2 x_{0}^{3}+\frac{1}{2} x_{0}=x_{0}^{2}+\frac{1}{4}, \text { hence } 2 x_{0}\left(x_{0}^{2}+\frac{1}{4}\right)=x_{0}^{2}+\frac{1}{4}
$$

Dividing by $x_{0}^{2}+\frac{1}{4}>0$, we get $2 x_{0}=1, x_{0}=\frac{1}{2}$. Thus, the coordinates of the point of tangency are $\left(\frac{1}{2} ; \frac{1}{4}\right)$ in the first quadrant; and $\left(-\frac{1}{2} ; \frac{1}{4}\right)$ in the second quadrant. Substituting these values into (1), the equations of the two tangents are: $y=x-\frac{1}{4}$, and $y=-x-\frac{1}{4}$.

Note. The problem can also be solved differently, for example, by determining the coordinates of the intersection point $T$ of the tangent $e$ and the $x$-axis, and applying the cosine rule to the triangle $F E T$, from which the value of $x_{0}$ can be calculated.

![](https://cdn.mathpix.com/cropped/2024_05_02_2b06091fd7d1dd6d32dcg-1.jpg?height=599&width=461&top_left_y=2091&top_left_x=821)