$P$ 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 at 14:56
  • $\begingroup$ @NickD that doesn’t give us $c$ $\endgroup$ – Aditya Jun 14 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.

| cite | improve this answer | |
  • $\begingroup$ How did you obtain (1)? And how did you get the equation for PS? $\endgroup$ – Aditya Jun 15 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 at 14:26
  • $\begingroup$ I don’t think PQS are collinear. PST and QSR are collinear $\endgroup$ – Aditya Jun 15 at 16:03
  • $\begingroup$ @Aditya Right, I meant to say that. $\endgroup$ – Tavish Jun 15 at 16:28
  • $\begingroup$ Solving the quadratic gives two answers, how do you which is the right one? $\endgroup$ – Aditya Jun 15 at 17:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.