11.4. On the coordinate plane, the parabola $y=x^{2}$ is drawn. On the positive $O y$-axis, a point $A$ is taken, and through it, two lines with positive slopes are drawn. Let $M_{1}, N_{1}$ and $M_{2}, N_{2}$ be the points of intersection with the parabola of the first and second line, respectively. Find the ordinate of point $A$, given that $\angle M_{1} O N_{1}=\angle M_{2} O N_{2}$, where $O$ is the origin.
Answer. 1. Solution. Let $a$ be the ordinate of point A. The line passing through point $A$ has the equation $\mathrm{y}=k \cdot x+a$, and the abscissas $x_{1}, x_{2}$ of points M and $N$ of intersection of the line with the parabola are the roots of the equation $x^{2}=k \cdot x+a$. From Vieta's theorem, we have $x_{1} \cdot x_{2}=-a$ and $x_{1}+x_{2}=k$. We calculate $\cos (\angle M O N)=$ $\frac{(\overrightarrow{M O}, \overrightarrow{N O})}{|M \hat{O}||\overrightarrow{O O}|}=\frac{x 1 \cdot x 2 \cdot(1+x 1 \cdot x 2)}{|x 1 \cdot x 2| \sqrt{\left(1+x 1^{2}\right)\left(1+x 2^{2}\right)}}=\frac{-a(1-a)}{a \sqrt{1+a^{2}-2 a+k^{2}}}=\frac{a-1}{\sqrt{1+a^{2}-2 a+k^{2}}}$. When $a=1$, we have $\cos (\angle M O N)=0$ and, therefore, $\angle M O N$ does not depend on $k$. If $a \neq 1$ and for two angular coefficients $k_{1}, k_{2}>0$ the corresponding angles are equal, then from the equality $\frac{a-1}{\sqrt{1+a^{2}-2 a+k 1^{2}}}=\frac{a-1}{\sqrt{1+a^{2}-2 a+k 2^{2}}}$ it follows that $k_{1}^{2}=k_{2}^{2}$, i.e., $k_{1}=$ $k_{2}$ (due to the positivity of $k_{1}, k_{2}$). Thus, we obtain the answer $a=1$. (The equality $k_{1}=k_{2}$ can also be obtained without using the dot product: instead, the tangent of the angle $M O N$ can be calculated using the tangent addition formula for angles MOA and NOA.)