Prove that the area of triangle formed by the tangents at $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ to the parabola $y^{2} = 4 a x (a > 0)$ is $\frac{1}{16 a} | (y_{1} - y_{2}) (y_{2} - y_{3}) (y_{3} - y_{1}) |$ sq. units.

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Solution

Let $(a {t_{1}}^{2}, 2 a t_{1}), (a {t_{2}}^{2}, 2 a t_{2})$ and $(a {t_{3}}^{2}, 2 a t_{3})$ be the three points on parabola $y_{2} = 4 a x$ such that,
$(x_{1}, y_{1}) = (a {t_{1}}^{2}, 2 a t_{1})$
$(x_{2}, y_{2}) = (a {t_{2}}^{2}, 2 a t_{2})$
$(x_{3}, y_{3}) = (a {t_{3}}^{2}, 2 a t_{3})$
The equation of tangents at $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ are respectively-
$t_{1} y = x + a {t_{1}}^{2} . . . . . (1)$
$t_{2} y = x + a {t_{2}}^{2} . . . . . (2)$
$t_{3} y = x + a {t_{3}}^{2} . . . . . (3)$
Subtracting equation $(2)$ from $(1)$ , we get
$y = a (t_{1} + t_{2})$
Substituting the value of $y$ in equation $(1)$ , we get
$a t_{1} t_{2}$
The point of intersection of the tangents at $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $(a t_{1} t_{2}, a (t_{1} + t_{2}))$ .
Similarly the points of intersection of tangent $(x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is $(a t_{2} t_{3}, a (t_{2} + t_{3}))$ and of $(x_{3}, y_{3})$ and $(x_{1}, y_{1})$ is $(a t_{1} t_{3}, a (t_{1} + t_{3}))$ .
Therefore area of triangle formed by the points of intersection of tangents is-
$A = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} a t_{1} t_{2} & a (t_{1} + t_{2}) & 1 a t_{2} t_{3} & a (t_{2} + t_{3}) & 1 a t_{1} t_{3} & a (t_{1} + t_{3}) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$\Rightarrow A = \frac{a^{2}}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} t_{1} t_{2} & (t_{1} + t_{2}) & 1 t_{2} t_{3} & (t_{2} + t_{3}) & 1 t_{1} t_{3} & (t_{1} + t_{3}) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$\Rightarrow A = \frac{a^{2}}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} t_{1} (t_{2} - t_{3}) & (t_{2} - t_{3}) & 1 t_{2} (t_{2} - t_{1}) & (t_{2} - t_{1}) & 1 t_{1} t_{3} & (t_{1} + t_{3}) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$\Rightarrow A = \frac{a^{2}}{2} (t_{2} - t_{3}) (t_{2} - t_{1}) ∣ ∣ ∣ ∣ \begin{matrix} t_{1} & 1 & 0 t_{3} & 1 & 0 t_{1} t_{3} & t_{1} + t_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣$
$\Rightarrow A = \frac{a^{2}}{2} | (t_{2} - t_{3}) (t_{2} - t_{1}) (t_{3} - t_{1}) |$
$\Rightarrow A = \frac{a^{2}}{16} | 2 a (t_{2} - t_{3}) 2 a (t_{2} - t_{1}) 2 a (t_{3} - t_{1}) |$
$\Rightarrow A = | (y_{1} - y_{2}) (y_{2} - y_{3}) (y_{3} - y_{1}) |$
Hence proved that the area of triangle formed by the points of intersection of tangents at given points to the parabola $y^{2} = 4 a x$ is $| (y_{1} - y_{2}) (y_{2} - y_{3}) (y_{3} - y_{1}) |$

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