Two perpendicular tangents to the parabola y^ | vedclass

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Question

(B) The equation of the tangent to the parabola $y^2 = 4ax$ at point $t$ is given by $ty = x + at^2$.Let the two tangents be at points $t_1$ and $t_2$

Vedclass · Accepted Answer

$x + a = 0$

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(B) The equation of the tangent to the parabola $y^2 = 4ax$ at point $t$ is given by $ty = x + at^2$.Let the two tangents be at points $t_1$ and $t_2$. Their equations are $t_1y = x + at_1^2$ and $t_2y = x + at_2^2$.The slopes of these tangents are $m_1 = \frac{1}{t_1}$ and $m_2 = \frac{1}{t_2}$.Since the tangents are perpendicular,$m_1 	imes m_2 = -1$,which implies $\frac{1}{t_1} 	imes \frac{1}{t_2} = -1$,so $t_1t_2 = -1$.The point of intersection of these two tangents is $(at_1t_2, a(t_1 + t_2))$.Substituting $t_1t_2 = -1$,the point of intersection becomes $(-a, a(t_1 + t_2))$.Since the $x$-coordinate is always $-a$,the locus of the intersection point is the line $x = -a$,which can be written as $x + a = 0$.

Two perpendicular tangents to the parabola $y^2 = 4ax$ always intersect on the line:

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