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If ($ x_1$,$ y_1$) , ($ x_2$,$ y_2$) & ($ x_3$,$ y_3$) be three points on the parabola $y^2= 4ax$ and the normals at these points meet in a point then prove that $\frac{ x_1 - x_2}{y_3} + \frac{ x_2 - x_3}{y_1} + \frac{ x_3 - x_1}{y_2}$=0.

Normal Equation:

$y=mx-am^3-2am$

Let (x',y') be common points

We get:

$y'=mx'-am^3-2am$

Let $m_1, \ m_2$ & $m_3$ be the slopes at ($ x_1$,$ y_1$) , ($ x_2$,$ y_2$) & ($ x_3$,$ y_3$).

We get

$ \ y_1=\ m_1 \ x_1+y'-\ m_1x'$

$ \ y_2=\ m_2 \ x_2+y'-\ m_2x'$

$ \ y_3=\ m_3 \ x_3+y'-\ m_3x'$

To arrive at the desired result i used

$y'=mx'-am^3-2am$

After this step i used $ \ m_1+ \ m_2+ \ m_3=0$ as $\ m^2$ coefficient is '0' but to no avail

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  • $\begingroup$ @Aretino I have posted a parabola problem where normal to three point intersect at one point, i showed the steps i have done but got struck $\endgroup$ – Samar Imam Zaidi Aug 30 '17 at 6:43
  • $\begingroup$ The points from where there are three normals to the parabola must reside on the $x$-axis, mustn't they? $\endgroup$ – Michael Hoppe Aug 30 '17 at 6:57
  • $\begingroup$ @MichaelHoppe The three points are on the parabola. $\endgroup$ – amd Aug 30 '17 at 6:58
  • $\begingroup$ @amd Of course, but if their normals meet in $P$, then $P$ must live on the first axis. $\endgroup$ – Michael Hoppe Aug 30 '17 at 7:06
  • $\begingroup$ @MichaelHoppe Must it? With $a=1$, the normals through $(4,4)$, $(3,-2\sqrt3)$ and $(7-4\sqrt3,-4+2\sqrt3)$ meet at $(9 - 2 \sqrt3, -6 + 4 \sqrt3)$. $\endgroup$ – amd Aug 30 '17 at 7:31
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Taking a slightly different approach from yours, we have $\nabla(y^2-4ax)=(-4a,2y)^T$, so in homogeneous coordinates the normal through a point $[x:y:1]$ that lies on the parabola is $\mathbf n=[y:2a:-(x+2a)\,y]$. For the normals through three points to have a common intersection, their scalar triple product $\mathbf n_1\times\mathbf n_2\cdot\mathbf n_3$ must vanish, therefore $$\begin{vmatrix} y_1 & 2a & -(x_1+2a)\,y_1 \\ y_2 & 2a & -(x_2+2a)\,y_2 \\ y_3 & 2a & -(x_3+2a)\,y_3 \end{vmatrix} = -2a((x_1-x_2)y_1y_2 + (x_2-x_3)y_2y_3 + (x_3-x_1)y_1y_3)=0$$ which is equivalent to the desired condition if $y_1,y_2,y_3\ne0$.

To put this in terms with which you might be more familiar, if you negate the last column of the above matrix, you have the augmented coefficient matrix of the system of linear equations of the three normals. This system is overdetermined, so for it to have a solution, the rows of the matrix must be linearly dependent, which is equivalent to the determinant of the matrix being zero.

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  • $\begingroup$ Thanks for your prompt answer $\endgroup$ – Samar Imam Zaidi Aug 30 '17 at 8:18
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Using $4ax_i = y_i^2$ we can rewrite the statement to be proved as

$\displaystyle\sum_{cyc} y_1y_2(y_1^2-y_2^2) = (y_1-y_2)(y_2-y_3)(y_3-y_1)(y_1+y_2+y_3)$

The parametric form of the Normal at $(at^2,2at)$ is $y+xt = at^3+2at$. If it passes through $(h,k)$, we must have $at^3+(2a-h)t-k=0$

From this we see that if $t_1,t_2,t_3$ are the roots of this equation $\sum t_i=0 \Rightarrow \sum y_i =0$. With this the required statement is readily seen to be true.

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