Parametric Equations and the Eccentric Angle

Why parametrise the ellipse

The equation $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ describes the ellipse but doesn't hand you individual points on it conveniently — solving for $y$ given $x$ involves a square root, and picking points scattered evenly around the curve by trial is clumsy. A parametric form, where both $x$ and $y$ are written as functions of a single angle, fixes this and turns out to connect the ellipse to something very familiar: a circle.

The auxiliary circle and the eccentric angle

Draw the circle of radius $a$ centred at $C$, using the major axis $AA'$ as its diameter — this is called the auxiliary circle of the ellipse, with equation $x^2+y^2=a^2$. Since $b<a$, the ellipse sits entirely inside this circle, touching it only at $A$ and $A'$.

Now take any point $P$ on the ellipse. Draw a perpendicular from $P$ to the major axis and extend it to meet the auxiliary circle at $Q$. The angle $\theta = \angle ACQ$ (measured at the centre, from $CA$ to $CQ$) is called the eccentric angle of $P$. As $P$ travels once around the ellipse, $\theta$ sweeps through the full range $0$ to $2\pi$.

Since $Q$ lies on the circle of radius $a$, its coordinates are $(a\cos\theta, a\sin\theta)$ — so $x = a\cos\theta$ for the point $P$ too, because $P$ and $Q$ share the same foot on the major axis. Substituting $x=a\cos\theta$ into the ellipse equation:

Taking the branch consistent with $Q$'s position gives the parametric equations of the ellipse:

For brevity, the point $(a\cos\theta, b\sin\theta)$ is often just called "the point $\theta$" and written $P(\theta)$. Notice these two equations together are exactly equivalent to the single Cartesian equation — plugging $x=a\cos\theta$, $y=b\sin\theta$ into $x^2/a^2+y^2/b^2$ gives $\cos^2\theta+\sin^2\theta=1$ for every $\theta$, automatically.

This parametric form is what makes later results — the equations of the tangent and normal at a point — so clean, because a single parameter $\theta$ replaces the pair $(x_1,y_1)$ subject to the constraint of lying on the ellipse.

Worked Example

Problem. Find the point on the ellipse $\dfrac{x^2}{49}+\dfrac{y^2}{9}=1$ whose eccentric angle is $\theta = 120°$. Then, in the other direction, find the eccentric angle of the point $\left(\dfrac{7}{2}, \dfrac{3\sqrt3}{2}\right)$ on the same ellipse.

Solution. Here $a=7$, $b=3$. …