Parametric Equations of the Circle, Parabola, Ellipse and Hyperbola

(i) Circle $x^2+y^2=a^2$. Let $P(x,y)$ be a point on the circle, $OP$ making angle $\theta$ with the $x$-axis, and $M$ the foot of the perpendicular from $P$ to the $x$-axis. From right triangle $OPM$: $x=OM=a\cos\theta$, $y=MP=a\sin\theta$. So $x=a\cos\theta,\ y=a\sin\theta$ ($0\le\theta\le2\pi$) parametrise the circle; conversely, squaring and adding recovers $x^2+y^2=a^2\cos^2\theta+a^2\sin^2\theta=a^2$.

(ii) Parabola $y^2=4ax$. For $P(x_1,y_1)$ on the parabola, $y_1^2=4ax_1$. Writing $\dfrac{y_1}{2a}=t$ (an arbitrary real number, since $y_1$ ranges over all reals), $y_1=2at$ and then $x_1=\dfrac{y_1^2}{4a}=\dfrac{(2at)^2}{4a}=at^2$. So $x=at^2,\ y=2at$ ($-\infty<t<\infty$) parametrise the parabola; eliminating $t$ (i.e. $t=y/2a$) recovers $y^2=4ax$.

(iii) Ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$. Let $P$ be on the ellipse and $Q$ the corresponding point directly above/below it on the auxiliary circle $x^2+y^2=a^2$, with $\angle ACQ=\alpha$ (so $Q=(a\cos\alpha,a\sin\alpha)$). Since $P$ shares its $x$-coordinate with $Q$, $x=a\cos\alpha$; substituting into the ellipse equation and solving for $y$ gives $y=b\sin\alpha$. So $P=(a\cos\alpha,b\sin\alpha)$; the parameter $\alpha$ (or $\theta$) is called the eccentric angle of $P$ — note carefully that $\theta$ is the angle $CQ$ makes with the $x$-axis, not the angle $CP$ makes with it. So $x=a\cos\theta,\ y=b\sin\theta$ ($0\le\theta\le2\pi$) parametrise the ellipse.

(iv) Hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$. By an entirely analogous construction (using $\sec^2\theta-\tan^2\theta=1$ in place of $\cos^2\theta+\sin^2\theta=1$), $x=a\sec\theta,\ y=b\tan\theta$ parametrise the hyperbola, for $-\pi\le\theta\le\pi$, $\theta\ne\pm\pi/2$.

Reference table.