Normal with a Given Slope

Normal to x²/a² + y²/b² = 1 with slope m. Up to four normals can be drawn from a point inside the evolute of the ellipse.

The normal at eccentric angle $\theta$ has slope $(a/b)\tan\theta$. Setting this equal to $m$:

$$\tan\theta = \frac{bm}{a} \implies \sin\theta = \frac{bm}{\sqrt{a^2+b^2m^2}}, \quad \cos\theta = \frac{a}{\sqrt{a^2+b^2m^2}}$$

Substituting into the parametric normal $ax\sec\theta - by\csc\theta = a^2-b^2$:

$$ax \cdot \frac{\sqrt{a^2+b^2m^2}}{a} - by \cdot \frac{\sqrt{a^2+b^2m^2}}{bm} = a^2-b^2$$ $$x\sqrt{a^2+b^2m^2} - \frac{y\sqrt{a^2+b^2m^2}}{m} = a^2-b^2$$ $$x - \frac{y}{m} = \frac{a^2-b^2}{\sqrt{a^2+b^2m^2}}$$ $$y = mx - \frac{m(a^2-b^2)}{\sqrt{a^2+b^2m^2}}$$

Point of contact:

$$\left(\frac{a^2}{\sqrt{a^2+b^2m^2}},\; \frac{-b^2m}{\sqrt{a^2+b^2m^2}}\right)$$

Note: for each slope $m$, there is one normal (as with the parabola) — contrast with tangents, where two exist per slope.