Parametrization of Implicitly Defined Functions

Unit 4 · Functions Involving Parameters, Vectors, and Matrices

What AP Precalc asks here

Ellipses parametrize naturally with cosine and sine because $\cos^2 + \sin^2 = 1$ . Hyperbolas use secant and tangent because $\sec^2 - \tan^2 = 1$ . Once the parametrization is chosen, you can verify it by substituting into the implicit equation — the trig identity does the rest. The chosen $t$ -interval determines whether you trace the whole curve, a single branch, or just a piece.

Standard parametrizations

Ellipse centered at

(h, k)

x = h + a \cos t, \;\; y = k + b \sin t

Hyperbola (right branch)

x = h + a \sec t, \;\; y = k + b \tan t

AP Tip: The parametrization

x = h + a \sec t, y = k + b \tan t

traces only one branch of the hyperbola at a time. The other branch needs

x = h - a \sec t

Type 1

Parametrize an ellipse

Use $x = h + a \cos t, y = k + b \sin t$ .

Example 1

Parametrize

\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1

. (A)

x = 9 \cos t, y = 4 \sin t

(B)

x = 3 \cos t, y = 2 \sin t

(C)

x = \cos(3t), y = \sin(2t)

(D)

x = 3 \sin t, y = 2 \cos t

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Type 2

Parametrize a hyperbola

Use $x = h + a \sec t, y = k + b \tan t$ for one branch.

Example 2

Parametrize one branch of

\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1

. (A)

x = 2 \cos t, y = 3 \sin t

(B)

x = 4 \sec t, y = 9 \tan t

(C)

x = 2 \sec t, y = 3 \tan t

(D)

x = 2 \sin t, y = 3 \cos t

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Type 3

Verify a parametrization

Substitute into the implicit equation. The trig identity should reduce both sides to the same value.

Example 3

Verify that

x = 2 \cos t, y = 3 \sin t

lies on

\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1

. (A) Yes (B) No — produces

\cos^2 t + \sin^2 t = 4/9

t = 0

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