Parametrizing Implicit Curves — AP® Precalculus Notes

Beyond the exam. The AP® Precalculus exam does not test this unit. Many classes teach it, and it prepares you for calculus.

You can now describe curves two ways: as an equation in $x$ and $y$ (an implicitly defined curve) and as a parametric function $(x(t), y(t))$ . This page turns the first kind into the second, and back. Turning an equation into parametric equations is called finding a parametrization.

Graphs of functions

The easiest case: if the curve is $y = f(x)$ , let $x$ itself be the parameter.

x(t) = t, \qquad y(t) = f(t).

For $y = x^2 - 3x$ , use $(t, t^2 - 3t)$ ; it traces the parabola from left to right as $t$ increases. If the curve is $x = g(y)$ , a function of $y$ , use $y$ as the parameter instead: $x = y^2 + 1$ becomes $(t^2 + 1, t)$ , traced from bottom to top.

Ellipses: the Pythagorean identity

For $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2}$ $= 1$ , the two fractions are squares that add to $1$ , just like $\cos^2 t + \sin^2 t = 1$ . So set

\frac{x - h}{a} = \cos t, \quad \frac{y - k}{b} = \sin t \qquad\Rightarrow\qquad x(t) = h + a\cos t, \quad y(t) = k + b\sin t.

Worked example 1. Parametrize $\frac{(x - 1)^2}{9} + \frac{(y - 2)^2}{4}$ $= 1$ .

$h = 1$ , $k = 2$ , $a = 3$ , $b = 2$ : $x(t) = 1 + 3\cos t$ , $y(t) = 2 + 2\sin t$ , $0 \le t \le 2\pi$ .

Check: $\frac{(3\cos t)^2}{9}$ $+ \frac{(2\sin t)^2}{4}$ $= \cos^2 t$ $+ \sin^2 t$ $= 1$ . The trip starts at $(4, 2)$ when $t = 0$ and goes once around counterclockwise.

Figure 1.

(3\cos t, 2\sin t)

: the circle of radius

3

squeezed vertically by the factor

\frac{2}{3}

. LevlPrep original diagram.

As with circles, $k - b\sin t$ reverses the direction, and $\cos(2t)$ and $\sin(2t)$ go around twice as fast. The parameter $t$ is not the angle of the point from the center unless $a = b$ .

Hyperbolas: the other Pythagorean identity

Divide $\sin^2 t + \cos^2 t = 1$ by $\cos^2 t$ to get $\tan^2 t + 1 = \sec^2 t$ , so $\sec^2 t - \tan^2 t = 1$ . A difference of squares equal to $1$ is exactly the hyperbola's form:

x(t) = h + a\sec t, \qquad y(t) = k + b\tan t.

Worked example 2. Parametrize $\frac{x^2}{4} - \frac{y^2}{9} = 1$ and say which branch $-\frac{\pi}{2} < t < \frac{\pi}{2}$ gives.

$x(t) = 2\sec t$ , $y(t) = 3\tan t$ . Check: $\frac{4\sec^2 t}{4} - \frac{9\tan^2 t}{9}$ $= \sec^2 t - \tan^2 t$ $= 1$ .

For $-\frac{\pi}{2} < t < \frac{\pi}{2}$ , $\cos t > 0$ , so $\sec t \ge 1$ and $x \ge 2$ : the right branch. As $t$ increases, $\tan t$ increases, so the point moves up the branch, from far below to far above. The values $t = \pm\frac{\pi}{2}$ are left out, where $\sec t$ and $\tan t$ are undefined; $\frac{\pi}{2} < t < \frac{3\pi}{2}$ gives the left branch.

Going back: removing the parameter

To find the equation a parametric curve satisfies, solve each coordinate for the trig function and use the matching identity. For $x = 1 + 3\cos t$ , $y = 2 + 2\sin t$ : $\cos t = \frac{x - 1}{3}$ and $\sin t = \frac{y - 2}{2}$ , so $\left(\frac{x - 1}{3}\right)^2 + \left(\frac{y - 2}{2}\right)^2$ $= 1$ . For polynomials, solve one coordinate for $t$ and substitute: $(t^2 - 1, t)$ gives $t = y$ and $x = y^2 - 1$ .

Many parametrizations, one curve

Curve	A parametrization	Identity used
$y = f(x)$	$(t, f(t))$	none
$x = g(y)$	$(g(t), t)$	none
ellipse	$(h + a\cos t, k + b\sin t)$	$\cos^2 t + \sin^2 t = 1$
hyperbola	$(h + a\sec t, k + b\tan t)$	$\sec^2 t - \tan^2 t = 1$

None of these is the only choice. $(t^3, t^6)$ also traces all of $y = x^2$ , because $t^3$ takes every real value. But $(t^2, t^4)$ traces only the right half, since $t^2 \ge 0$ . Always check that the parameter interval covers the whole curve you want, and nothing more.