Parametrics question

[Aviator_13]

The distinct points P(2ap,ap²) and Q(2aq,aq²)  lie on the parabola x² = 4ay . The points P and Q are constrained to move such that QR ⟂ PR where R is the fixed point (2a,a), which also lies on the parabola, given pq + p + q = -5. By drawing the diagram, or otherwise, find the values of p and q for which the relationship QR ⟂ PR is not possible.

EDIT:
After like a shi* ton of trial and error on desmos i think the answer might be smth like p and q cannot equal to plus/minus 1.

[RuiAce]

\[ \text{Presumably you've already set }m_{QR}m_{PR}=-1\\ \text{to obtain that result }pq+p+q=-5. \]
\[ \text{Now, notice that rearranging this gives}\\ \begin{align*}p(q+1) &= -(5+q)\\ p &= -\frac{q+5}{q+1}.\end{align*} \]
\[ \text{From here, the condition that }\boxed{q\neq -1}\text{ should be clear}\\ \text{because otherwise }p\text{ is undefined.}\\ \text{And then, by doing the computations in an alternate way,}\\ \text{it can also be shown that }\boxed{p\neq-1}\]
As for the restriction that \(p,q\neq +1\), I think that was kinda hinted in the question. Because if \(p=1\), then \(P\) would coincide with \(R\), so \(PR\) is just a single point and not a line segment. (Same goes for if \(q=1\) for \(Q\).)

[jamonwindeyer]

Hey! Welcome to the forums!

I was curious about this too, but I agree with Rui, I think \(p\neq1\) and \(q\neq1\) is.because either of those conditions makes Point P or Q the same as Point R, which doesn't make sense in terms of the question. So those are out!

For the other condition, you can actually stop earlier in your working, I do this:




This rearranged gives the equation you are given in the question, but it also shows the other issue - \(p\neq-1\) and \(q\neq-1\), because if they are, this equation becomes \(0=-1\) - Also nonsense!

[Aviator_13]

Thanks a lot!!