When do these normals to a parabola meet on the parabola?

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Taking the derivative of the equation \(y^2=4ax\) with respect to \(x\) we find that \[2y\frac{dy}{dx}=4a \qquad \Longrightarrow \qquad \frac{dy}{dx}=\frac{2a}{y}=\frac{2a}{2at}=\frac{1}{t}.\] The gradient of normal to the parabola at the point \((at^2,2at)\) is therefore \(-t\) since \[\frac{1}{t}\times -t=-1.\] The equation of the normal is then \[y-2at=-t(x-at^2) \qquad \Longrightarrow \qquad y=-tx+2at+at^3.\]

The gradient of the line \(PQ\) is \[\frac{2at_2-2at_1}{at_2^2-at_1^1}=\frac{2(t_2-t_1)}{(t_2-t_1)(t_1+t_2)}=\frac{2}{t_1+t_2}.\] The equation of the line through points \(P\) and \(Q\) is then \[y-2at_1=\frac{2}{t_1+t_2}(x-at_1^2).\] This will pass through the point \((-2a,0)\) if and only if \[\begin{align*} 0-2at_1&=\frac{2}{t_1+t_2}(-2a-at_1^2) \\ &\iff t_1 = \frac{1}{t_1+t_2}(2+t_1^2) \\ &\iff t_1(t_1+t_2) = 2+t_1^2 \\ &\iff t_1 t_2 = 2, \end{align*}\]

as required.

The two normals passing through \(P\) and \(Q\) have the equations \[\begin{align*} y &=-t_1x+2at_1+at_1^3 \\ y &=-t_2x+2at_2+at_2^3, \end{align*}\] using the first part of the question. They intersect when \[\begin{align*} -t_1x+2at_1+at_1^3 &= -t_2x+2at_2+at_2^3 \\ \Longrightarrow \qquad 2a(t_2-t_1)+a(t_2^3-t_1^3) &= x(t_2-t_1) \\ \end{align*}\] and factorising \((t_2^3-t_1^3)\) gives \[\begin{align*} \qquad 2a(t_2-t_1)+a(t_2-t_1)(t_1^2+t_1t_2+t_2^2) &= x(t_2-t_1) \end{align*}\] Since \(t_1\neq t_2\), as we are assuming \(P\) and \(Q\) are distinct, we can divide through by \((t_2-t_1)\) to get \[x=a(t_1^2+t_1t_2+t_2^2+2).\] Substituting for \(2=t_1t_2\) we find that \[\begin{align*} x &=a(t_1^2+2t_1t_2+t_2^2) \\ &= a(t_1+t_2)^2. \end{align*}\] Using this expression, we can substitute back into one of the original equations for the normal to find \[\begin{align*} y &= -t_1a(t_1+t_2)^2+2at_1+at_1^3 \\ &=-t_1^3a-2at_1^2t_2-at_1t_2^2+2at_1+at_1^3\\ &=-2a \times 2 \times t_1 - 2 \times at_2+2at_1\\ &= -2a(t_1+t_2). \end{align*}\]

Thus \(R=\left(a(t_1+t_2)^2,-2a(t_1+t_2)\right)\), and comparing this to the general point on the parabola \((at^2,2at)\), we can see that \(R\) is clearly on the curve.

Since \(P = (at_1^2, 2at_1)\), \(Q = (at_2^2, 2at_2)\), \(R = (a(t_1+t_2)^2, -2a(t_1+t_2))\), and \(O = (0, 0)\), then:

\(OP\) has gradient \(\dfrac{2}{t_1}\), \(PQ\) has gradient \(\dfrac{2}{t_1+t_2}\), \(QR\) has gradient \(-\dfrac{2}{t_1}\), while \(RO\) has gradient \(-\dfrac{2}{t_1+t_2}\).

Thus \(OP\) and \(QR\) are reflections of each other in a vertical or horizontal line, and \(PQ\) and \(OR\) are likewise.

The diagram shows that angle \(\widehat{PQR}\) is \(180^\circ-\alpha -\beta\), and so the angles \(\widehat{POR}\) and \(\widehat{PQR}\) add to \(180^\circ\), and therefore \(OPQR\) is a cyclic quadrilateral.

Thus \(O\) lies on the circumcircle of \(PQR\), as required.