Prove that the area of the triangle formed by the normals to the parabola at the points $(a t_{1}^{2}, 2 a t_{1}), (a t_{2}^{2}, 2 a t_{2})$ and $(a t_{3}^{2}, 2 a t_{3})$ is $\frac{a^{2}}{2} (t_{2} - t_{3}) (t_{3} - t_{1}) (t_{1} - t_{2}) (t_{1} + t_{2} + t_{3})^{2} .$

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Solution

The normal of the point $(a t^{2}, 2 a t)$ is $y + x t = 2 a t + a t^{3}$
$N_{1} : y + x t_{1} = 2 a t_{1} + a t_{1}^{3}$ at point $(a t_{1}^{2}, 2 a t_{1})$
$N_{2} = y + x t_{2} = 2 a t_{2} + a t_{2}^{3}$ at point $(a t_{2}^{2}, 2 a t_{2})$
$N_{3} = y + x t_{3} = 2 a t_{3} + a t_{3}^{3}$ at point $(a t_{3}^{2}, 2 a t_{3})$
The triangle $A B C$ formed by the tangents as its side equations.
point $A : (x_{1}, y_{2}) = (2 a + a (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2}), - a t_{1} t_{2} (t_{1} + t_{2}))$
point $B : (x_{2}, y_{2}) = (2 a + a (t_{2}^{2} + t_{3}^{2} + t_{2} t_{3}), - a t_{2} t_{3} (t_{2} + t_{3}))$
point $C : (x_{3}, y_{3}) = (2 a + a (t_{1}^{2} + t_{3}^{2} + t_{1} t_{3}), - a t_{1} t_{3} (t_{1} + t_{3})$
The Area of the triangle $A B C$ is given by
$\frac{1}{2} (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}))$
substituting the values of $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , $(x_{3}, y_{3})$
Evaluate $x_{1} (y_{2} - y_{3}) = (2 a + a (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2}) (- a t_{2} t_{3} (t_{2} + t_{3}) - (- a t_{1} t_{3} (t_{1} + t_{3})) = (2 a + a (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2})) (t_{1} - t_{2}) a t_{3} (t_{1} + t_{2} + t_{3})$
similarly $x_{2} (y_{3} - y_{1})$ = $(2 a + a (t_{2}^{2} + t_{3}^{2} + t_{2} t_{3})) (t_{2} - t_{3}) a t_{1} (t_{1} + t_{2} + t_{3})$
$x_{3} (y_{1} - y_{2}))$ = $(2 a + a (t_{1}^{2} + t_{3}^{2} + t_{1} t_{3}) (t_{3} - t_{1}) a t_{2} (t_{1} + t_{2} + t_{3})$
Therefore Add three of them, we get
The area of the triangle as $\frac{a^{2}}{2} (t_{2} - t_{3}) (t_{3} - t_{1}) (t_{1} - t_{2}) (t_{1} + t_{2} + t_{3})^{2}$

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