$P$ 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$... $$P$$ 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? • Presumably, you have to use the given fact that PQ passes through$(-2, 3)$, Jun 14 '20 at 14:56 • @NickD that doesn’t give us$c$Jun 14 '20 at 15:20 1 Answer 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. • How did you obtain (1)? And how did you get the equation for PS? Jun 15 '20 at 7:29 • @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 $$(y-y_1) = \frac{y_2-y_1}{x_2-x_1} (x-x_1)$$ Jun 15 '20 at 14:26
• I don’t think PQS are collinear. PST and QSR are collinear Jun 15 '20 at 16:03
• @Aditya Right, I meant to say that. Jun 15 '20 at 16:28
• Solving the quadratic gives two answers, how do you which is the right one? Jun 15 '20 at 17:03