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

I tried as follows.

Let the normals meet at $(h,k)$. Then, $$am^3 + (2a-h)m + k = 0.$$

After Solving the equation, $$ x_1 y_1(y_2-y_3) + x_2 y_2(y_3-y_1) + x_3 y_3 (y1 - y2) = 0 $$

Substituting $x_k = y_k^2/4a$ for $k \in \{1,2,3\}$

But I am not getting the result.

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The equation of the normal at $(am^2,2am)$

$$mx+y-(2am+am^3)=0$$

Now if the normal passes through $(h,k)$ $$mh+k-(2am+am^3)=0\iff am^3-m(2a+h)-k=0$$

Its roots are $t_i$s where $1\le i\le3$ and $x_i=at_i^2,y_i=2at_i$

$\implies\sum t_i=0$

$$\dfrac{x_1-x_2}{y_3}=\dfrac{t^2_1-t_2^2}{2t_3}=\dfrac{t_2-t_1}2$$

Can you take it from here?

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  • $\begingroup$ How would you eliminate unknown $h$ & $k$ from cubic equation? @Lab $\endgroup$ – Bhaskara-III Dec 29 '15 at 13:42
  • $\begingroup$ @Bhaskara-III, Vieta's formula says : $$\sum t_i=\dfrac0a$$ $\endgroup$ – lab bhattacharjee Dec 29 '15 at 13:43

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