$TP$ 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.


1 Answer 1


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.


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.