**Topic**

Let the parabola (*P*) has the canonical equation: *y*^{2} = 2*px* (*p* > 0) and the line *x* = *m* (*m *> 0) cut (*P*) at two points *I*, *KY* distinguish. Prove two points *I* and *KY* symmetric about the axis *Ox*.

**Solution method – See details**

Step 1: Parameterizing coordinates *I*, *KY* straight line PT *x *= *m*

Step 2: Change the coordinates *I*, *KY* into PT (*P*) and prove that the coordinates of these two points have opposite signs and then conclude

**Detailed explanation**

Since \(I,K \in d:x = m\) \(I(m;t),K(m;k)\)

Since \(I,K \in (P)\) \(\left\{ \begin{array}{l}{t^2} = 2pm\\{k^2} = 2pm\end{array} \ right.\)\( \Leftrightarrow {t^2} = {k^2} \Leftrightarrow \left\{ \begin{array}{l}t = k\\t = – k\end{array} \right. \)

With *t* = *k* then *I* and *KY* duplicate \( \Rightarrow \) *t* = *k* unsatisfactory

With *t* = –*k* then *I*(*m *; *t*) and *KY*(*m *; *-t*). Then *I* and *KY *symmetric about the axis *Ox* (PCM)