AP® Precalculus review sheet from Aim for Five (aimforfive.com/precalc/units/4/4-7)
Unit 4 · Topic 4.7
4.7 Parametrization of Implicitly Defined Functions
Many implicitly defined curves can be written as parametric functions. Graphs of functions and their inverses parametrize easily, ellipses use cosine and sine, and hyperbolas use secant and tangent, all because they satisfy the original equation for every t.
Key terms
- parametrization
- ellipse parametrization
- hyperbola parametrization
- Pythagorean identity
What makes a parametrization work
A parametrization (x(t), y(t)) works for an implicit curve if substituting x(t) for x and y(t) for y makes the equation true for every t in the domain. That's how you check one.
A curve has many correct parametrizations. They may differ in direction, starting point and speed.
Functions and inverses
Any function y = f(x) can be parametrized by letting x be the parameter: (x(t), y(t)) = (t, f(t)).
If f is invertible, its inverse can be parametrized by swapping: (x(t), y(t)) = (f(t), t), over an appropriate interval of t. No algebra is needed to find the inverse formula.
The same idea works for parabolas. If the equation can be solved for y in terms of x, use (t, f(t)). If it can be solved for x in terms of y, use (f(t), t).
Example: y = eˣ is parametrized by (t, eᵗ), and its inverse, y = ln x, by (eᵗ, t), for all real t.
Replacing t with −t reverses the direction a curve is traced, which is handy when a problem asks for a particular direction.
Ellipses
The ellipse (x − h)²/a² + (y − k)²/b² = 1 is traced by x(t) = h + a cos t, y(t) = k + b sin t for 0 ≤ t ≤ 2π.
Check: ((a cos t)²)/a² + ((b sin t)²)/b² = cos² t + sin² t = 1. A circle is the case a = b.
Hyperbolas
The hyperbola (x − h)²/a² − (y − k)²/b² = 1, opening left and right, is traced by x(t) = h + a sec t, y(t) = k + b tan t for 0 ≤ t ≤ 2π, skipping t = π/2 and t = 3π/2, where sec t and tan t are undefined.
The hyperbola (y − k)²/b² − (x − h)²/a² = 1, opening up and down, is traced by x(t) = h + a tan t, y(t) = k + b sec t over the same values of t.
These work because of the identity sec² t − tan² t = 1, which comes from 1 + tan² t = sec² t in topic 3.12. For the left-right hyperbola, t-values where sec t > 0 trace the right branch and those where sec t < 0 trace the left branch.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Parametrizing an ellipse
Parametrize (x − 2)²/9 + (y + 3)²/4 = 1 and verify it.
Show the solutionHide the solution
- Step 1: Here h = 2, k = −3, a = 3, b = 2.
- Step 2: x(t) = 2 + 3 cos t, y(t) = −3 + 2 sin t, for 0 ≤ t ≤ 2π.
- Step 3: Verify: (3 cos t)²/9 + (2 sin t)²/4 = cos² t + sin² t = 1.
- Step 4: At t = 0 the point is (5, −3), the rightmost point, and it moves counterclockwise.
Answer: x(t) = 2 + 3 cos t, y(t) = −3 + 2 sin t, 0 ≤ t ≤ 2π.
- Example 2
Parametrizing a hyperbola
Parametrize (x − 1)²/4 − y²/9 = 1.
Show the solutionHide the solution
- Step 1: It opens left and right, with h = 1, k = 0, a = 2, b = 3.
- Step 2: x(t) = 1 + 2 sec t, y(t) = 3 tan t, for 0 ≤ t ≤ 2π with t ≠ π/2, 3π/2.
- Step 3: Verify: (2 sec t)²/4 − (3 tan t)²/9 = sec² t − tan² t = 1.
- Step 4: At t = 0: (1 + 2, 0) = (3, 0), the right vertex.
Answer: x(t) = 1 + 2 sec t, y(t) = 3 tan t, with t ≠ π/2, 3π/2.
- Example 3
Trap: choosing the parameter for a sideways parabola
Parametrize the parabola x = (y − 2)² + 1.
Show the solutionHide the solution
- Step 1: Setting x = t would force you to solve t = (y − 2)² + 1 for y, which gives two branches, y = 2 ± √(t − 1). That's messy and misses the point.
- Step 2: The equation is already solved for x, so let y be the parameter: y(t) = t and x(t) = (t − 2)² + 1.
- Step 3: This traces the whole parabola, from bottom to top as t increases.
Answer: x(t) = (t − 2)² + 1, y(t) = t, for all real t.
Common mistakes
- Using cos t and sin t for a hyperbola. A hyperbola needs sec t and tan t, because of the minus sign.
- Swapping a and b, so the ellipse is stretched the wrong way. a goes with x, b with y.
- Including t = π/2 or 3π/2 in a secant-tangent parametrization, where both are undefined.
On the exam
- Unit 4 is not on the AP Precalculus Exam, so you'll see this topic on class tests rather than in May.
- Expect to parametrize a conic from its equation and to verify a parametrization by substituting it into the equation.
Connected topics
Videos
Check yourself
3 questions on 4.7 Parametrization of Implicitly Defined Functions. Pick an answer to see if you got it, and why.
Which of the following parametrizes the ellipse (x − 1)²/4 + (y + 2)²/25 = 1 for 0 ≤ t ≤ 2π?
Which of the following parametrizes the hyperbola x²/9 − y²/4 = 1?
Which of the following parametrizations traces the ellipse x²/9 + y² = 1 exactly once clockwise, starting at (0, 1), for 0 ≤ t ≤ 2π?
0 of 3 answered