Let A_1 and A_2 be points on the axis of the parabola y²

Let $A_{1}$ and $A_{2}$ be points on the axis of the parabola $y^{2} = 2px$ at equal distances from the focus, and let $d_{1},d_{2}$ be their distances to a tangent of the parabola. Show that $d_{1}^{2}-d_{2}^{2}$ is constant.

Step 1: The setting

The focus is $F = \left(\dfrac{p}{2},0\right)$ and the axis is $Ox$. With $r\gt 0$ the common distance, $A_{1} = \left(\frac{p}{2}+r,\,0\right), \qquad A_{2} = \left(\frac{p}{2}-r,\,0\right)$.

Step 2: The tangent

At $\bigl(x_{0},y_{0}\bigr)$ on the parabola the tangent is $y_{0}y = p\bigl(x+x_{0}\bigr)$ (Problem 496), i.e.

\[ px-y_{0}y+px_{0} = 0. \tag{1} \]

Step 3: The two distances

For a point $\bigl(u,0\bigr)$ on the axis, (1) gives $d = \dfrac{p\bigl|u+x_{0}\bigr|}{\sqrt{p^{2}+y_{0}^{2}}}$, so

\[ d_{1}^{2}-d_{2}^{2} = \frac{p^{2}}{p^{2}+y_{0}^{2}}\left[\left(\frac{p}{2}+x_{0}+r\right)^{2}-\left(\frac{p}{2}+x_{0}-r\right)^{2}\right]. \tag{2} \]

Step 4: Simplify

By $\bigl(A+r\bigr)^{2}-\bigl(A-r\bigr)^{2} = 4Ar$ with $A = \dfrac{p}{2}+x_{0}$, $d_{1}^{2}-d_{2}^{2} = \frac{4p^{2}r\left(\dfrac{p}{2}+x_{0}\right)}{p^{2}+y_{0}^{2}}$.

Now the equation of the curve enters: $y_{0}^{2} = 2px_{0}$, hence

\[ p^{2}+y_{0}^{2} = p^{2}+2px_{0} = 2p\left(\frac{p}{2}+x_{0}\right), \] and the bracket cancels completely: \[d_{1}^{2}-d_{2}^{2} = \frac{4p^{2}r}{2p} = 2pr. \tag{3}\]

Step 5: Verification

\[p = 2,\ r = 1:\ F = \bigl(1,0\bigr),\ A_{1} = \bigl(2,0\bigr),\ A_{2} = \bigl(0,0\bigr),\ \text{and } 2pr = 4\]

\[\text{Tangent at } \bigl(1,2\bigr):\ x-y+1 = 0:\quad d_{1}^{2} = \frac{9}{2},\ d_{2}^{2} = \frac12 \;\Longrightarrow\; d_{1}^{2}-d_{2}^{2} = 4 \;\;\checkmark\]

\[\text{Tangent at } \bigl(4,4\bigr):\ x-2y+4 = 0:\quad d_{1}^{2} = \frac{36}{5},\ d_{2}^{2} = \frac{16}{5} \;\Longrightarrow\; d_{1}^{2}-d_{2}^{2} = 4 \;\;\checkmark\]