11.2. The tangent to the graph of $y=x^{2}$ intersects the $O x$ and $O y$ axes at points $A$ and $B$. Find the lengths of the legs $O A$ and $O B$, if the area of triangle $A O B$ is 16.
Answer: $O A=2, O B=16$. Solution. Let $\left(x_{0}, x_{0}^{2}\right)$ be the point of tangency, for definiteness $x_{0}>0$ (if $x_{0}<0$, the tangent line intercepts segments of the same length on the axes as at ( $-x_{0}$ )). The equation of the tangent line is $y-x_{0}^{2}=2 x_{0}\left(x-x_{0}\right)$, (since $\left.\left(x^{2}\right)^{\prime}=2 x\right)$. From this, we obtain the coordinates of the points of intersection with the axes: $A\left(\frac{x_{0}}{2}, 0\right), B\left(0,-x_{0}^{2}\right)$. Then $S=\frac{1}{2} \cdot \frac{x_{0}}{2} \cdot x_{0}^{2}=\frac{x_{0}^{3}}{4}=16 \Rightarrow x_{0}=4$, from which the result follows: $O A=2, O B=16$.