The chord $x-y-1=0$ cuts the parabola $y^2-4x=0$ at P and Q (1st and 4th quad respectively). Normals at P and Q meet at R

Question

Find the point slope of normal through R and point of concurrency of normals through P,Q and R

I found $P(3+2\sqrt 2, 2+2\sqrt 2)$ and $Q(3-2\sqrt 2, 2-2\sqrt 2)$.

If I sovle usinf the standard procedure, I can find the values of $R$ by finding the two normal equations and solving them. But that’s a very lengthy process. I want to save my calculations. Is there any property that can help me?

I don’t need a solution, I just need methods to save time.

Thanks a lot

"Find the point slope of normal through R and point of concurrency of normals through P,Q and R" is not clear to me. — Intelligenti pauca, May 28 '20 at 18:07
$PQ$ is a focal chord, hence tangents at $P$ and $Q$ are perpendicular and meet at $R'$ on the directrix. $R$ is then the reflection of $R'$ about line $PQ$. — Intelligenti pauca, May 28 '20 at 21:49

score 0 · Answer 1 · answered May 29 '20 at 08:19

Methods to save time.

As $PQ$ is a focal chord, then we know that tangents at at $P$ and $Q$ are perpendicular and meet at $R′$ on the directrix $x=-1$.
From the resolvent equation $y^2=4y+4$ we can obtain without solving the midpoint $M=(3,2)$ of $PQ$.
Line $R'M$ is parallel to the axis, hence $R'=(-1,2)$.
$M$ is the midpoint of $RR'$, hence $R=(7,2)$.

For the final part I see no alternative to the standard path of writing the equation of the generic normal through $(a^2,2a)$, impose it passes through $R$ and factor away solutions $P$ and $Q$ to find a third solution.

The chord $x-y-1=0$ cuts the parabola $y^2-4x=0$ at P and Q (1st and 4th quad respectively). Normals at P and Q meet at R

1 Answers1