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Two perpendicular tangents to $y^2=4 a x$ always intersect on the line, if
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$x+a=0$
We know that tangent to the parabola at points $t_1$ and $t_2$ are $t_1 y=x+a t_1^2$ and $t_2 y=x+a t_2^2$. Since
tangents are perpendicular to the parabola, therefore, $\frac{1}{t_1} \cdot \frac{1}{t_2}=-1$ or $t_1 t_2=-1$
We also know that their point of intersection $=\left(a t_1 t_2, a\left(t_1+t_2\right)\right)=\left(-a, a\left(t_1+t_2\right)\right)$
Thus these points lie on directrix $x=-a$ or $x+a=0$
tangents are perpendicular to the parabola, therefore, $\frac{1}{t_1} \cdot \frac{1}{t_2}=-1$ or $t_1 t_2=-1$
We also know that their point of intersection $=\left(a t_1 t_2, a\left(t_1+t_2\right)\right)=\left(-a, a\left(t_1+t_2\right)\right)$
Thus these points lie on directrix $x=-a$ or $x+a=0$
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