$P$ and $Q$ are two points on the parabola $y^2=8x$ and $S$ is the focus. $PS$ and $QS$ meet the curve at $T$and $R$. If $PQ$ passes through a fixed point $(-2,3)$, then find the fixed point through which $TR$ passes

Let the parameters for $P,Q,T,R$ be $t_1,t_2,t_3,t_4$ respectively.

So $t_1t_3=t_2t_4=-1$

The equation for PQ will be $$y=\frac{1}{t_1+t_2}x+c$$ where $c$ is unknown.

If I could find $c$, it would be easy to compare with the line $TR$ and use the relationship mentioned before. But I couldn’t.

How should I solve it?

  • $\begingroup$ Presumably, you have to use the given fact that PQ passes through $(-2, 3)$, $\endgroup$
    – NickD
    Jun 14 '20 at 14:56
  • $\begingroup$ @NickD that doesn’t give us $c$ $\endgroup$
    – Aditya
    Jun 14 '20 at 15:20

You have the parametric points $P(2p^2,4p)$ and $Q(2q^2,4q)$. Since $PQ$ passes through $(-2,3)$, we have the relation $$ \frac{4p-3}{2p^2+2}=\frac{4q-3}{2q^2+2} \implies 3(p+q)+4-4pq=0 \hspace{1 cm}\mathbf{ (1)}$$ The equation of $PS$ can be calculated to be $$x=y\left(\frac{2p}{p^2-1}\right) + 2$$ Satisfy this with $y^2=8x$ to get the quadratic $$py^2-y(4p^2-4)-16p=0$$ Either by using the quadratic formula or by dividing by $y-4p$, the coordinates of $T$ can be obtained: $T=\left( \frac{2}{p^2},\frac{-4}{p}\right) $

And similarly $R=\left( \frac{2}{q^2},\frac{-4}{q}\right) $. Then the equation if $TR$ is $$ (p+q)y +(2pq)x + 4=0$$ Looking carefully at $\mathbf{(1)}$ should suggest that setting $y=3$ and $x=-2$ will always satisfy the above equation.

  • $\begingroup$ How did you obtain (1)? And how did you get the equation for PS? $\endgroup$
    – Aditya
    Jun 15 '20 at 7:29
  • $\begingroup$ @Aditya $P$, $Q$ and $S$ are colinear, so the slope of $PS$ equals that of $QS$. Also, the equation of a line passing through two points is $$(y-y_1) = \frac{y_2-y_1}{x_2-x_1} (x-x_1) $$ $\endgroup$
    – Tavish
    Jun 15 '20 at 14:26
  • $\begingroup$ I don’t think PQS are collinear. PST and QSR are collinear $\endgroup$
    – Aditya
    Jun 15 '20 at 16:03
  • $\begingroup$ @Aditya Right, I meant to say that. $\endgroup$
    – Tavish
    Jun 15 '20 at 16:28
  • $\begingroup$ Solving the quadratic gives two answers, how do you which is the right one? $\endgroup$
    – Aditya
    Jun 15 '20 at 17:03

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