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$TP$ and $TQ$ are tangents to the parabola $y^2=4ax$, and the normals at $P$ and $Q$ meet at a point $R$ on the curve. Prove that the center of the circle circumscribing the triangle $TPQ$ lies on the parabola $2y^2=a(x-a)$.


This is a solved example in my book. I could not understand few lines.

Let the parabola be $y^2=4ax$ and the coordinates of $P$ and $Q$ be $(at^2_1,-2at_1)$ and $(at^2_2,-2at_2)$, respectively. The equations of the tangent lines are $$\begin{align} \text{at}\;P: &\quad t_1y=x+at^2_1 \\ \text{at}\;Q: &\quad t_2y=x+at^2_2 \end{align}$$ Solving these two, the coordinates of point of intersection is $T(at_1t_2,a(t_1+t_2))$.

Now, the equation of the normal at $P$ is $$y=-t_1x+2at_1+at^3_1$$ Let this normal pass through $R(at^2_3,-2at_3)$ on the parabola. Then $$t_1+\frac{2}{t_1}=-t_3 \tag{1}$$ Similarly, if the normal at $Q$ passes through $R$, then $$t_2+\frac{2}{t_2}=-t_3 \tag{2}$$ So, $$t_1+\frac{2}{t_1}=t_2+\frac{2}{t_2}=-t_3$$ Subtracting we get, $$(t_1-t_2)\left(1-\frac{2}{t_1t_2}\right)=0$$ So, since $t_1 \neq t_2$, $$t_1t_2=2 \tag{3}$$ Again, let the center of the circle passing through $P$, $Q$, $T$ be $(x,y)$, we have $$2x=a(t_1+t_2)^2+2a \tag{4}$$ and $$2y=a(t_1+t_2)(1-t_1t_2) \tag{5}$$ Now, eliminate $t_1$ and $t_2$ from (4) and (5), using $t_1t_2=2$,we get $$2y^2=a(x-a)$$

I could not understand how they got equation $(4)$ and $(5)$. How did they obtain the circumcenter of $\triangle TPQ$? The rest of the answer, I understood.

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There does appear to be a bit of a leap (there's also an apparent typo), but the reasoning turns out to be pretty simple.

The typo is that $R$ should be $(at_3^2, 2at_3)$ (a positive $2$ in the $y$-coordinate). This is because the normals at $P$ and $Q$ (with a negative $2$) must meet on the "other" side of the parabola. Now ...


$\triangle TPR$ and $\triangle TQR$ are right triangles with common hypotenuse $\overline{TR}$. The midpoint of that hypotenuse is the circumcenter of those triangles, which then must also be the circumcenter of $\triangle TPQ$ (and, in fact, $\square PRQT$). So, the circumcenter is $$\frac12(T+R) = \frac{a}{2} \left(\; t_1 t_2 + t_3^2, t_1 + t_2 + 2 t_3 \;\right) \tag{$\star$}$$ Now, because $t_1t_2 = 2$, we have $2/t_1 = t_2$, so that (from (1) or (2)) $t_3 = -(t_1+t_2)$. Consequently, $(\star)$ becomes $$\frac{a}{2}\left(\; 2 + (t_1+t_2)^2,\;-(t_1 + t_2)\;\right) \tag{$\star\star$}$$ and $$x-a = \frac{a}{2}(t_1+t_2)^2 = \frac{2}{a}y^2 \qquad\to\qquad 2y^2 = a(x-a)$$


Remark. Since the solution mentions invoking $t_1t_2=2$ only after the circumcenter calculation, it seems that the author had a different approach in mind. Maybe the author determined the perpendicular bisectors of $\overline{PT}$ and $\overline{QT}$, and then found their intersection; but that's a lot of algebra to skip over. shrug

Remark 2. That circumcircle also includes the vertex of the original parabola. (Proof left as an exercise to the reader.) Nifty.

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