Unit 7 · Topic 4.7 · Optional

Parametrization of Implicitly Defined Functions

Replace one equation relating x and y with two coordinate functions of a shared parameter, then verify the curve, its coverage, and its direction.

Learning Goals

Verify a parametrization by substitution into an implicit equation.
Parameterize equations solved for x or y by choosing the other coordinate as t.
Build trigonometric parametrizations of circles, ellipses, and hyperbolas.
Select a parameter interval that traces the intended branch or portion.
Distinguish algebraic validity from complete coverage of a curve.

1. From an Implicit Relation to a Parametric Curve

An implicit equation \(F(x,y)=0\) describes all ordered pairs that satisfy one condition. A parametrization assigns both coordinates to one input:

\[x=x(t),\qquad y=y(t).\]

It is valid on a stated domain when

\[F\bigl(x(t),y(t)\bigr)=0\]

for every allowed t. As t changes, the point \((x(t),y(t))\) traces the curve.

2. Validity and Coverage Are Different

Question	How to answer it
Validity	Substitute \(x(t)\) and \(y(t)\) into the original equation.
Coverage	Use the parameter domain and coordinate ranges to decide which points are reached.
Direction	Evaluate points at increasing t-values.
Repetition	Check whether different t-values produce the same point.

A formula can generate only correct points and still omit part of the intended graph.

3. The Direct Choice \(x=t\)

If the relation is solved as \(y=f(x)\), choose

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

The domain of t must match the x-values allowed in the original relation. This simple choice preserves the graph while adding a direction from smaller to larger x.

4. Parameterize a Vertical Parabola

For \(y=x^2-3\), set \(x=t\). Then

\[x(t)=t,\qquad y(t)=t^2-3,\qquad -\infty<t<\infty.\]

Substitution gives \(y(t)=x(t)^2-3\), so every generated point lies on the parabola. Because every real x-value occurs as \(t\), the entire parabola is covered.

If the requested curve is only the right half, use \(t\ge0\).

5. The Direct Choice \(y=t\)

If the equation is solved as \(x=g(y)\), choose \(y=t\) and \(x=g(t)\). For the sideways parabola \(x=y^2-1\),

\[x(t)=t^2-1,\qquad y(t)=t,\qquad t\in\mathbb R.\]

Substitution gives \(x(t)=y(t)^2-1\). This method avoids splitting the relation into \(y=\pm\sqrt{x+1}\), so both halves are traced continuously.

6. Trigonometric Identities Build Closed Curves

The identity \(\cos^2t+\sin^2t=1\) guarantees every parameterized point lies on the ellipse.

The identity

\[\cos^2t+\sin^2t=1\]

matches the sum of two squared terms in a circle or ellipse. Horizontal and vertical scale factors stretch the unit circle into the required dimensions.

7. Circle Parametrization

For \((x-h)^2+(y-k)^2=r^2\), use

\[x(t)=h+r\cos t,\qquad y(t)=k+r\sin t.\]

On \(0\le t\le2\pi\), the circle is traced once counterclockwise, beginning and ending at \((h+r,k)\).

8. Verify the Circle Algebraically

\[\begin{aligned}(x(t)-h)^2+(y(t)-k)^2 &=r^2\cos^2t+r^2\sin^2t\\ &=r^2(\cos^2t+\sin^2t)\\ &=r^2.\end{aligned}\]

The identity proves validity for every real t. The interval \(0\le t\le2\pi\) supplies one complete traversal.

9. Ellipse Parametrization

For an axis-aligned ellipse

\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1,\]

match each semiaxis to its coordinate:

\[x(t)=h+a\cos t,\qquad y(t)=k+b\sin t,\qquad 0\le t\le2\pi.\]

After substitution, the denominators cancel the squared scale factors and leave \(\cos^2t+\sin^2t=1\).

10. Worked Example: Parameterize an Ellipse

Parameterize \(\frac{x^2}{9}+\frac{y^2}{4}=1\). The horizontal semiaxis is 3 and the vertical semiaxis is 2, so

\[x(t)=3\cos t,\qquad y(t)=2\sin t,\qquad 0\le t\le2\pi.\]

\[\frac{(3\cos t)^2}{9}+\frac{(2\sin t)^2}{4}=\cos^2t+\sin^2t=1.\]

t	\((x(t),y(t))\)	Location
\(0\)	\((3,0)\)	Right vertex
\(\pi/2\)	\((0,2)\)	Top co-vertex
\(\pi\)	\((-3,0)\)	Left vertex
\(3\pi/2\)	\((0,-2)\)	Bottom co-vertex
\(2\pi\)	\((3,0)\)	Return to start

11. Control Portion and Direction

t-domain for \(x=a\cos t,\ y=b\sin t\)	Trace
\([0,2\pi]\)	Full ellipse, counterclockwise
\([0,\pi]\)	Upper half, right to left
\([-\pi/2,\pi/2]\)	Right half, bottom to top
\([0,4\pi]\)	Full ellipse twice

Replacing t by \(-t\) reverses the direction while leaving the geometric curve unchanged.

12. Horizontal Hyperbolas Use Secant and Tangent

The identity \(\sec^2t-\tan^2t=1\) matches

\[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1.\]

A parametrization is

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

Substitution produces \(\sec^2t-\tan^2t=1\). Use \(-\frac{\pi}{2}<t<\frac{\pi}{2}\) for the right branch and \(\frac{\pi}{2}<t<\frac{3\pi}{2}\) for the left branch. Values \(t=\frac{\pi}{2}+k\pi\) are excluded because secant and tangent are undefined.

13. Vertical Hyperbolas Swap the Roles

For

\[\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1,\]

place secant with the positive squared term:

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

The interval \(-\frac{\pi}{2}<t<\frac{\pi}{2}\) traces the upper branch because \(\sec t>0\); the interval \(\frac{\pi}{2}<t<\frac{3\pi}{2}\) traces the lower branch because \(\sec t<0\).

14. Conic Parametrization Summary

Curve	Parametrization	Identity or method
\(y=f(x)\)	\(x=t,\ y=f(t)\)	Direct substitution
Ellipse	\(x=h+a\cos t,\ y=k+b\sin t\)	\(\cos^2t+\sin^2t=1\)
Horizontal hyperbola	\(x=h+a\sec t,\ y=k+b\tan t\)	\(\sec^2t-\tan^2t=1\)
Vertical hyperbola	\(x=h+a\tan t,\ y=k+b\sec t\)	\(\sec^2t-\tan^2t=1\)

15. AP Workflow and Common Errors

Identify the implicit curve and put it in standard form when needed.
If one variable is isolated, choose the other coordinate as t.
For an ellipse or hyperbola, match the equation to the correct identity.
Attach every translation and scale factor to the correct coordinate.
Substitute both coordinate functions into the original equation.
State a t-domain and determine the portion, direction, and repetition.
Check excluded parameter values, especially for secant and tangent.

Do not stop after proposing formulas; verify them.
Do not interchange the horizontal and vertical semiaxes.
Do not use \(\cos t,\sin t\) for a difference-of-squares equation.
Do not include parameter values where secant or tangent is undefined.
Do not claim complete coverage from substitution alone.

Key Takeaways

Create parameterizations for implicit curves using trigonometric or algebraic identities.
Core relationship: \(x=a\cos t,\quad y=b\sin t\)
Error check: Match each semiaxis with the correct coordinate.

Checkpoint · Topic 4.7

Parameterize \(y=2x^2+1\) using \(x=t\), and verify by substitution.
Parameterize \(x=(y-2)^2+3\) without solving for two y-branches.
Give a parametrization and one-traversal domain for \(\frac{(x-1)^2}{16}+\frac{(y+2)^2}{9}=1\).
Restrict that ellipse parametrization so only its upper half is traced from right to left.
Parameterize \(\frac{(x+2)^2}{25}-\frac{(y-1)^2}{4}=1\), state separate domains for its two branches, and verify the identity.