P.T the area of the triangle formed by three tangents at $(x_{1}, y_{1}) (x_{2}, y_{2}) and (x_{3}, y_{3})$ to the parabola y² = 4ax is $\frac{1}{16 a} |(y_{1} - y_{2}) (y_{2} - y_{3}) (y_{3} - y_{1})| sq. units$

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Detailed Solution

A = Point of intersection of tangents at $t_{1}, t_{2} \Rightarrow [({at}_{1} t_{2}, a (t_{1} + t_{2})]$
B = P.T of tangents at $t_{2}, t_{3} \Rightarrow [({at}_{2} t_{3}, a (t_{2} + t_{3})]$
C = P.T of tangents at $t_{3}, t_{1} \Rightarrow [({at}_{3} t_{1}, a (t_{3} + t_{1})]$
Area of triangle ABC
$\begin{array}{l} = \frac{1}{2} |\begin{array}{ll} {at}_{1} t_{2} - {at}_{1} t_{3} & a (t_{2} - t_{3}) \\ {at}_{1} t_{3} - {at}_{2} t_{3} & a (t_{1} - t_{2}) \end{array}| \\ = \frac{1}{2} |\begin{array}{ll} {at}_{1} (t_{2} - t_{3}) a (t_{2} - t_{3}) \\ {at}_{3} (t_{1} - t_{2}) a (t_{1} - t_{2}) \end{array}| \\ = \frac{1}{2} |a (t_{2} - t_{3}) a (t_{1} - t_{2})| |\begin{array}{ll} t_{1} & 1 \\ t_{3} & 1 \end{array}| \\ = \frac{a^{2}}{2} |(t_{1} - t_{2}) (t_{2} - t_{3}) (t_{1} - t_{3})| \\ = \frac{a^{2}}{2} |(t_{1} - t_{2}) (t_{2} - t_{3}) (t_{3} - t_{1})| \\ = \frac{a^{2}}{2} |(\frac{y_{1} - y_{2}}{2 a}) (\frac{y_{2} - y_{3}}{2 a}) (\frac{y_{3} - y_{1}}{2 a})| \\ = \frac{1}{16 a} |(y_{1} - y_{2}) (y_{2} - y_{3}) (y_{3} - y_{1})| \end{array}$