AP Precalculus 4.7: Parametrization of Implicitly Defined Functions

In 4.5 Implicitly Defined Functions you met equations such as $y^2=x+4$ that are not solved for either variable. A parametrization of the curve writes both coordinates in terms of one new variable, the parameter $t$, as $(x(t),y(t))$.

The test is substitution. Put $x(t)$ in for $x$ and $y(t)$ in for $y$. The equation has to come out true for every value of $t$ in the domain, not for one or two of them.

Try $(x(t),y(t))=(t^2-4,t)$ on $y^2=x+4$. The right side becomes $(t^2-4)+4=t^2$, and the left side is $t^2$. The two sides are the same expression in $t$, so every point this produces is on the curve.

Now try $(t,t+2)$. At $t=0$ it gives $(0,2)$, and $2^2=0+4$ is true. At $t=1$ it gives $(1,3)$, and $3^2=1+4$ is false. One value of $t$ that works proves nothing. Substituting turns the equation into $t^2+4t+4=t+4$, which holds only at $t=0$ and $t=-3$.

Some curves hand you a parametrization. If $y=f(x)$, let $x$ be the parameter: $(x(t),y(t))=(t,f(t))$. Substituting gives $f(t)=f(t)$, true for every $t$.

The inverse of an invertible function swaps the roles: $(x(t),y(t))=(f(t),t)$, with $t$ running over the domain of $f$. That interval matters. Take $f(x)=x^2$ for $x\ge 0$. Its inverse is $(t^2,t)$ for $t\ge 0$, the top half of a parabola lying on its side. Let $t$ take every real value and you get the bottom half too, which is not the graph of the inverse.

The same idea works for any equation you can solve for one variable. A parabola that opens up or down can be solved for $y$, and then $x$ becomes $t$. One that opens left or right can be solved for $x$, and then $y$ becomes $t$.

The parabola $x-1=-2(y+3{)}^2$ from 4.6 Conic Sections opens left. Solve for $x$: $x=1-2(y+3{)}^2$. Replace $y$ with $t$:

At $t=-3$ that gives the vertex, $(1,-3)$.

An ellipse cannot be solved for $y$ as one function, so it needs a different tool: the identity ${\cos }^{2}t+{\sin }^{2}t=1$. The ellipse

is parametrized by $x(t)=h+a\cos t$ and $y(t)=k+b\sin t$ for $0\le t\le 2\pi$.

Watch it work on $\frac{(x-3{)}^2}{25}+\frac{(y+2{)}^2}{9}=1$, the left picture in the diagram. Here $h=3$, $k=-2$, $a=5$ and $b=3$, so $x(t)=3+5\cos t$ and $y(t)=-2+3\sin t$. Substitute. $x-3$ becomes $5\cos t$ and $y+2$ becomes $3\sin t$, so the left side is

for every $t$. At $t=0$ the point is $(8,-2)$, the rightmost point, and at $t=\frac{\pi }{2}$ it is $(3,1)$, the top.

The numbers in front of cosine and sine are the radii, 5 and 3, not the denominators 25 and 9. When $a=b$ the ellipse is a circle, and the same formulas give it.

A hyperbola uses another form of the Pythagorean identity, ${\tan }^{2}t={\sec }^{2}t-1$, which says ${\sec }^{2}t-{\tan }^{2}t=1$.

• Opening left and right, $\frac{(x-h{)}^2}{a^2}-\frac{(y-k{)}^2}{b^2}=1$ is $x(t)=h+a\sec t$, $y(t)=k+b\tan t$.
• Opening up and down, $\frac{(y-k{)}^2}{b^2}-\frac{(x-h{)}^2}{a^2}=1$ is $x(t)=h+a\tan t$, $y(t)=k+b\sec t$.

In both, $0\le t\le 2\pi$. Secant goes with the variable whose term is positive, because ${\sec }^{2}t$ is the positive term in ${\sec }^{2}t-{\tan }^{2}t=1$. Put the tangent there and the left side comes out as ${\tan }^{2}t-{\sec }^{2}t$, which is $-1$.

For the right picture in the diagram, $\frac{(x-2{)}^2}{9}-\frac{(y+1{)}^2}{16}=1$, that gives $x(t)=2+3\sec t$ and $y(t)=-1+4\tan t$. At $t=0$ the point is $(5,-1)$. At $t=\pi$, where $\sec t=-1$, it is $(-1,-1)$, on the other branch.

Secant and tangent do not exist where $\cos t=0$, so there is no point at $t=\frac{\pi }{2}$ or $t=\frac{3\pi }{2}$. As $t$ approaches either one, the point runs out along an asymptote.