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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.

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  • $\begingroup$ Welcome to Math.SE. This seems like a problem copied verbatim from somewhere, Please share your efforts in solving the problem so that someone can guide you. $\endgroup$ – Shailesh Dec 28 '15 at 13:55
  • $\begingroup$ Sorry but how to start $\endgroup$ – user101522 Dec 28 '15 at 13:56
  • $\begingroup$ In a standard parabola, you know the focus (a,0), directrix x + a = 0 and so on. The perpendicular lines pass through the focus.... What more do you need. Put all that together. $\endgroup$ – Shailesh Dec 28 '15 at 13:58
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    $\begingroup$ I showed what I have done $\endgroup$ – user101522 Dec 28 '15 at 14:07
  • $\begingroup$ Can you link or find relation between $k$ and $t?$ $\endgroup$ – Narasimham Dec 28 '15 at 14:52
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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.

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