Parametric Equations of the Hyperbola

Just as points on an ellipse can be written using a single angle parameter via its auxiliary circle, points on a hyperbola can be parametrised using the circle drawn on its transverse axis as diameter — called the auxiliary circle, $x^2+y^2=a^2$.

Here's the geometric idea: take a point $P(x,y)$ on the hyperbola and drop a perpendicular from it to the transverse axis, landing at $M$. Draw the tangent from $M$ to the auxiliary circle, touching it at $Q$, and let $\theta$ be the angle that $CQ$ makes with the transverse axis (where $C$ is the centre). Because $CQ=a$ is the radius and $CM$ is the hypotenuse of the right triangle $CQM$, we get $CM = a\sec\theta$ — and since $CM$ is just the $x$-coordinate of $P$, this gives $x = a\sec\theta$. Feeding this into the hyperbola's equation and solving for $y$ gives $y=\pm b\tan\theta$. Putting the two together, the standard parametric equations of the hyperbola are

(The two excluded angles are skipped simply because $\sec\theta$ blows up there — there's no finite point on the curve at those parameter values.) The point $(a\sec\theta,\,b\tan\theta)$ is often abbreviated as "the point $\theta$" or written $P(\theta)$.

This parametrisation is genuinely useful, not just decorative: because $\sec^2\theta - \tan^2\theta = 1$ is an identity that holds for every $\theta$, a point written this way automatically satisfies $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ without you needing to check it — which makes $\theta$ a convenient single free variable to sweep across the whole curve (both branches, since $\sec\theta$ takes both signs), instead of juggling $x$ and $y$ together.

Worked example. On $\dfrac{x^2}{9}-\dfrac{y^2}{4}=1$ ($a=3,b=2$), take $\theta = 60^\circ$. Then $\sec 60^\circ = 2$ and $\tan 60^\circ = \sqrt3$, so the point is …