Normal at a Point on the Ellipse

Normal to x²/a² + y²/b² = 1 at (x₁, y₁). Slope of normal = a²y₁/(b²x₁). The normal passes through neither focus in general.

At $P(x_1, y_1)$ on $x^2/a^2 + y^2/b^2 = 1$, the tangent slope is $-b^2x_1/(a^2y_1)$.

The normal is perpendicular, so its slope is $a^2y_1/(b^2x_1)$.

Normal through $P(x_1, y_1)$:

$$y - y_1 = \frac{a^2y_1}{b^2x_1}(x - x_1)$$ $$b^2x_1(y - y_1) = a^2y_1(x - x_1)$$ $$b^2x_1 y - b^2x_1y_1 = a^2y_1 x - a^2y_1x_1$$ $$a^2y_1 x - b^2x_1 y = a^2y_1x_1 - b^2x_1y_1 = x_1y_1(a^2-b^2)$$

Dividing by $x_1y_1$:

$$\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$$

Note: $a^2 - b^2 = c^2 = a^2e^2$ — the right-hand side depends only on the shape of the ellipse, not on the specific point.

The normal bisects the angle between the focal radii: This is the optical property of the ellipse — a ray from one focus reflects off the ellipse toward the other focus. The normal at any point bisects the angle $\angle S'PS$ (the angle between the two focal radii).