Show that the locus of the point of intersection of two tangents, which with the tangent at the vertex form a triangle of constant area $c^{2}$ , is the curve $x^{2} (y^{2} - 4 a x) = 4 c^{4} a^{2}$ .

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Solution

Let the equation of the parabola be $y^{2} = 4 a x$ , the two tangents be $t_{1} y = x + a t_{1}^{2}$ and $t_{2} y = x + a t_{2}^{2}$
The tangent at the vertex is $x = 0$
Thus, the three vertices of the formed triangle are $(a t_{1} t_{2}, a (t_{1} + t_{2})), (0, a t_{2}), (0, a t_{1})$
The area of the triangle can be written as $∣ ∣ ∣ \begin{matrix} 0 & 0 & a t_{1} t_{2} & 0 a t_{1} & a t_{2} & a (t_{1} + t_{2}) & a t_{1} \end{matrix} ∣ ∣ ∣ = \frac{1}{2} \times [(a t_{1} \times 0 + a t_{2} \times a t_{1} t_{2} + 0) - (0 \times a t_{2} + 0 \times a (t_{1} + t_{2}) + a t_{1} t_{2} \times a t_{1})]$
$∴ 2 c^{2} = a t_{1} t_{2} (t_{2} - t_{1})$
The point of intersection of tangents is given by $x = a t_{1} t_{2}, y = a (t_{1} + t_{2})$
$\Rightarrow 2 c^{2} = x \times \sqrt{\frac{y^{2}}{a^{2}} - \frac{4 x}{a}}$
$\Rightarrow 2 c^{2} a = x \times \sqrt{y^{2} - 4 a x}$
Squaring both sides, we have $4 c^{4} a^{2} = x^{2} (y^{2} - 4 a x)$ as the required locus.

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