### Instruction:
Let $T_{1}, T_{2}, T_{3}$ be points on a parabola, and $t_{1}, t_{2}, t_{3}$ the tangents to the parabola at these points. Compute the ratio of the area of triangle $T_{1} T_{2} T_{3}$ to the area of the triangle determined by the tangents.

### Response:

Choose a Cartesian system of coordinates such that the equation of the parabola is $y^{2}=4 p x$. The coordinates of the three points are $T_{i}\left(4 p \alpha_{i}^{2}, 4 p \alpha_{i}\right)$, for appropriately chosen $\alpha_{i}, i=1,2,3$. Recall that the equation of the tangent to the parabola at a point $\left(x_{0}, y_{0}\right)$ is $y y_{0}=2 p\left(x+x_{0}\right)$. In our situation the three tangents are given by$$ 2 \alpha_{i} y=x+4 p \alpha_{i}^{2}, \quad i=1,2,3 . $$If $P_{i j}$ is the intersection of $t_{i}$ and $t_{j}$, then its coordinates are $\left(4 p \alpha_{i} \alpha_{j}, 2 p\left(\alpha_{i}+\alpha_{j}\right)\right)$. The area of triangle $T_{1} T_{2} T_{3}$ is given by a Vandermonde determinant: $$ \pm \frac{1}{2}\left|\begin{array}{lll} 4 p \alpha_{1}^{2} & 4 p \alpha_{1} & 1 \\ 4 p \alpha_{2}^{2} & 4 p \alpha_{2} & 1 \\ 4 p \alpha_{3}^{2} & 4 p \alpha_{3} & 1 \end{array}\right|=\pm 8 p^{2}\left|\begin{array}{lll} \alpha_{1}^{2} & \alpha_{1} & 1 \\ \alpha_{2}^{2} & \alpha_{2} & 1 \\ \alpha_{3}^{2} & \alpha_{3} & 1 \end{array}\right|=8 p^{2}\left|\left(\alpha_{1}-\alpha_{2}\right)\left(\alpha_{1}-\alpha_{3}\right)\left(\alpha_{2}-\alpha_{3}\right)\right| . $$The area of the triangle $P_{12} P_{23} P_{31}$ is given by$$ \begin{aligned} & \pm \frac{1}{2}\left|\begin{array}{lll}4 p \alpha_{1} \alpha_{2} & 2 p\left(\alpha_{1}+\alpha_{2}\right) & 1 \\4 p \alpha_{2} \alpha_{3} & 2 p\left(\alpha_{2}+\alpha_{3}\right) & 1 \\4 p \alpha_{3} \alpha_{1} & 2 p\left(\alpha_{3}+\alpha_{1}\right) & 1\end{array}\right| \\ & =\pm 4 p^{2}\left|\begin{array}{ll}\alpha_{1} \alpha_{2}\left(\alpha_{1}+\alpha_{2}\right) & 1 \\\alpha_{2} \alpha_{3}\left(\alpha_{2}+\alpha_{3}\right) & 1 \\\alpha_{3} \alpha_{1}\left(\alpha_{3}+\alpha_{1}\right) & 1\end{array}\right|=\pm 4 p^{2}\left|\begin{array}{rr}\left(\alpha_{1}-\alpha_{3}\right) \alpha_{2}\left(\alpha_{1}-\alpha_{3}\right) & 0 \\\left(\alpha_{2}-\alpha_{1}\right) \alpha_{3}\left(\alpha_{2}-\alpha_{1}\right) & 0 \\\alpha_{3} \alpha_{1}\left(\alpha_{3}+\alpha_{1}\right) & 1\end{array}\right| \\ & =4 p^{2}\left|\left(\alpha_{1}-\alpha_{3}\right)\left(\alpha_{1}-\alpha_{2}\right)\left(\alpha_{2}-\alpha_{3}\right)\right| \text {. } \end{aligned} $$We conclude that the ratio of the two areas is 2 , regardless of the location of the three points or the shape of the parabola.