Tangent at Eccentric Angle θ

Tangent to the ellipse at the parametric point (a cosθ, b sinθ). Slope of this tangent is −(b cosθ)/(a sinθ) = −b cotθ/a.

Substitute $x_1 = a\cos\theta$, $y_1 = b\sin\theta$ into the point-form tangent $xx_1/a^2 + yy_1/b^2 = 1$:

$$\frac{x \cdot a\cos\theta}{a^2} + \frac{y \cdot b\sin\theta}{b^2} = 1$$ $$\frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1$$

Slope: Rewriting as $y = -\frac{b\cos\theta}{a\sin\theta} x + \frac{b}{\sin\theta}$, so slope $= -\frac{b\cos\theta}{a\sin\theta} = -\frac{b}{a}\cot\theta$.

Special cases:

• $\theta = 0$: tangent at $(a, 0)$ is $x/a = 1$, i.e. $x = a$ (vertical tangent at right vertex)
• $\theta = 90°$: tangent at $(0, b)$ is $y/b = 1$, i.e. $y = b$ (horizontal tangent at top vertex)

Chord joining two eccentric angles $\alpha$ and $\beta$:

$$\frac{x}{a}\cos\frac{\alpha+\beta}{2} + \frac{y}{b}\sin\frac{\alpha+\beta}{2} = \cos\frac{\alpha-\beta}{2}$$

When $\alpha \to \beta$, this becomes the tangent at $\theta = \alpha$:

$$\frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1 \checkmark$$