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If $t_{1}, t_{2}, t_{3}$ are three variable parameters satisfying the relations $a t_{2}^{2} t_{3}^{2} = 4 b (t_{2} + t_{3})$ and $a t_{2}^{2} t_{1}^{2} = 4 b (t_{1} + t_{3}),$ then the locus of the point of intersection of tangents to the parabola $y^{2} = 4 a x$ at $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{3}^{2}, 2 a t_{3})$ is

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$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

$\frac{y^{2}}{a^{2}} + \frac{x^{2}}{b^{2}} = 1$

$y^{2} = 4 b x$

$x^{2} = 4 b y$

answer is B.

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