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Let $(x_1,y_1), (x_2,y_2)$ and $(x_3,y_3)$ be three points on the parabola $y^2=4ax$, the normals at which meet in a point, then find $$\frac{x_1-x_2}{y_3}+\frac{x_2-x_3}{y_1}+\frac{x_3-x_1}{y_2}$$

My attempt: I tried putting it as $(at^2,2at)$ with $t_1,t_2$ and $t_3$ but I couldn't resolve it further. What else could I do?

ecrin
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Harshita Jain
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1 Answers1

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Denote by $(x,y)$ the meeting point of the three normals. Then, $$\forall i\in\{1,2,3\},\quad\frac{y-y_i}{x-x_i}=-\frac{y_i}{2a}$$ i.e. $$\forall i\in\{1,2,3\},\quad y_ix+2ay=2ay_i+x_iy_i,$$ from which we derive: $$\begin{align}0&=\begin{vmatrix}y_1&2a&2ay_1+x_1y_1\\y_2&2a&2ay_2+x_2y_2\\y_3&2a&2ay_3+x_3y_3\end{vmatrix}\\ &=2a\begin{vmatrix}y_1&1&x_1y_1\\y_2&1&x_2y_2\\y_3&1&x_3y_3\end{vmatrix}\\ &=2a\left(y_1y_2(x_1-x_2)+y_2y_3(x_2-x_3)+y_3y_1(x_3-x_1)\right),\end{align}$$ hence $$\frac{x_1-x_2}{y_3}+\frac{x_2-x_3}{y_1}+\frac{x_3-x_1}{y_2}=0.$$ If you haven't been taught determinants yet, here is an alternative proof: $$\begin{align}y_1y_2(x_1-x_2)&=y_2(x_1y_1)-y_1(x_2y_2)\\ &=y_2\left(y_1x+2a(y-y_1)\right)-y_1\left(y_2x+2a(y-y_2)\right)\\ &=2ay(y_2-y_1) \end{align}$$ and similarly, $$\begin{align}y_2y_3(x_2-x_3)&=2ay(y_3-y_2)\\ y_3y_1(x_3-x_1)&=2ay(y_1-y_3), \end{align}$$ and summing the three leads to the same conclusion.

Anne Bauval
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