Chord with a Given Midpoint

Equation of the chord of y² = 4ax whose midpoint is (x₁, y₁). Explicitly: yy₁ − 2a(x+x₁) = y₁² − 4ax₁.

Let $(x_1, y_1)$ be the midpoint of a chord of $y^2 = 4ax$ with endpoints at parameters $t_1$ and $t_2$.

Midpoint conditions:

$$\frac{at_1^2 + at_2^2}{2} = x_1, \quad \frac{2at_1 + 2at_2}{2} = y_1$$

From the second: $a(t_1 + t_2) = y_1$, so $t_1 + t_2 = y_1/a$.

Slope of the chord joining $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$:

$$m = \frac{2}{t_1 + t_2} = \frac{2}{y_1/a} = \frac{2a}{y_1}$$

Equation of the chord through $(x_1, y_1)$ with slope $2a/y_1$:

$$y - y_1 = \frac{2a}{y_1}(x - x_1)$$ $$y_1(y - y_1) = 2a(x - x_1)$$ $$yy_1 - y_1^2 = 2ax - 2ax_1$$ $$yy_1 - 2ax = y_1^2 - 2ax_1$$

In the shorthand notation $T = yy_1 - 2a(x+x_1)$ and $S_1 = y_1^2 - 4ax_1$:

$$T + 2ax_1 - 2ax_1 = S_1 + 4ax_1 - 2ax_1 \implies T = S_1$$

The chord with midpoint $(x_1, y_1)$ is $T = S_1$.