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If two perpendicular straight lines through the focus of the parabola $y^2 = 4ax$ meet its directrix in $T $ and $T'$ respectively. Show that the tangents to the parabola parallel to the perpendicular lines intersect in the mid point of $T T'$.
Progress. Let the points on parabola be $(at^2, 2at)$ and $(ak^2, 2ak)$ then the straight line through focus and these points meet directix at $(-a , -4at/(t^2 -1)]$ and $[-a ,-4ak/(k^2 -1)]$. Please give me a hint on how to continue.
Let $P(ap^2, 2ap)$ and $Q(aq^2,2aq)$ be two points on the parabola with $p\neq q$
The gradient of the tangent at $P$ is $\frac 1p$, so the equation of the tangent at $P$ is $$py=x+ap^2$$
Likewise the gradient of the tangent at $Q$ is $\frac 1q$, so the equation of the tangent at $Q$ is $$qy=x+aq^2$$
The tangents are perpendicular, so $pq=-1$. Furthermore they intersect at the point $T$ with coordinates $(apq, a(p+q))=(-a,a(p+q))$, confirming that the tangents intersect on the directrix.
Meanwhile, the two perpendicular lines parallel to the tangents and passing through the focus have equations $$py=x-a$$ and $$qy=x-a$$
These lines intersect the directrix at points with $y$ coordinates $-\frac{2a}{p}$ and $-\frac{2a}{q}$, whose midpoint is therefore $$(-a,-\frac{2a}{2}(\frac 1p+\frac 1q))=(-a, a(p+q))$$
Hence the result.
