Write the equation of the parabola and its derivative.

The parabola is given by $x^2 = 4ay$.
Differentiating with respect to $x$:

$$2x = 4a \frac{dy}{dx}$$ $$\frac{dy}{dx} = \frac{2x}{4a} = \frac{x}{2a}$$

This is the slope of the tangent at any point $(x_1, y_1)$ on the parabola.

The slope of the normal at $(x_1, y_1)$ is $m_N = -\frac{1}{\frac{dy}{dx}|_{(x_1, y_1)}} = -\frac{2a}{x_1}$.
The equation of the normal at $(x_1, y_1)$ is $y - y_1 = m_N (x - x_1)$.

$$y - y_1 = -\frac{2a}{x_1} (x - x_1)$$ $$x_1(y - y_1) = -2a(x - x_1)$$ $$x_1 y - x_1 y_1 = -2ax + 2ax_1$$ $$2ax + x_1 y - (x_1 y_1 + 2ax_1) = 0$$

Since $(x_1, y_1