Equation of the Normal to the Ellipse

From tangent slope to normal slope

The normal to the ellipse at a point $P$ is the line through $P$ perpendicular to the tangent there. From the tangent equation at $(x_1,y_1)$, $\dfrac{xx_1}{a^2}+\dfrac{yy_1}{b^2}=1$, rewriting as $y = -\dfrac{b^2x_1}{a^2y_1}x + (\text{const})$ shows the tangent's slope is $-\dfrac{b^2x_1}{a^2y_1}$. The normal, being perpendicular, has slope $\dfrac{a^2y_1}{b^2x_1}$ (negative reciprocal). Using the point-slope form through $(x_1,y_1)$ and simplifying:

Two edge cases fall outside this formula (since it divides by $x_1$ and $y_1$): if $x_1=0$ (P is an end of the minor axis), the normal is simply the y-axis; if $y_1=0$ (P is an end of the major axis), the normal is the x-axis — both are visually obvious once you picture the ellipse.

Normal in terms of the eccentric angle

Substituting $x_1=a\cos\theta$, $y_1=b\sin\theta$ into the normal equation and simplifying gives the normal at the point $\theta$:

(At those excluded angles the point sits on an axis, and the normal is just $y=0$ or $x=0$ as noted above.)

A note on how many normals pass through a point

Unlike the tangent — where a point on the ellipse has exactly one tangent — a point off the ellipse can have several normals drawn to the curve from it. Substituting a fixed point $(x_1,y_1)$ into the normal equation and rewriting $\cos\theta,\sin\theta$ in terms of $t=\tan(\theta/2)$ turns the condition into a quartic (degree-4) equation in $t$. A degree-4 polynomial has at most 4 real roots, so at most four normals can be drawn from any given point to an ellipse.

Worked Example

Problem. Find the equation of the normal to the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$ at the point where the eccentric angle is $\theta = 90°$, and separately at the point $(3, 16/5)$.

Solution. Here $a^2=25 \Rightarrow a=5$, $b^2=16\Rightarrow b=4$, so $a^2-b^2 = 9$. …