Prove that the locus of the point of intersection of two tangents which intercept a given distance 4c on the tangent at the vertex is an equal parabola.

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Solution

Tangent at the vertex of the parabola $y^{2} = 4 a x$ is $x = 0$
The intercept of the first tangent $t_{1} y = x + a t_{1}^{2}$ on the line $x = 0$ is given by $(0, a t_{1})$ , while that of the second tangent $t_{2} y = x + a t_{2}^{2}$ is given by $(0, a t_{2})$
$a (t_{1} - t_{2}) = 4 c$
The intersection of these tangents is given by $x = a t_{1} t_{2}, y = a (t_{1} + t_{2})$
$y^{2} = a^{2} (t_{1} + t_{2})^{2} = a^{2} (t_{1}^{2} + 2 t_{1} t_{2} + t_{2}^{2}) = a^{2} (t_{1} - t_{2})^{2} + 4 a^{2} t_{1} t_{2}$
$∴ y^{2} = 16 c^{2} + 4 a x$ is the required locus

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