What are the semi-axis coefficients for parametrizing (x^2)/(25) + (y^2)/(9) = 1? [1 point]

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How do you find a parametrization for an implicitly defined curve such as a circle or ellipse?

Topic 4.7 Parametrization of Implicitly Defined Functions: find parametric equations that trace an implicitly defined curve, and verify the parametrization satisfies the implicit equation.

A focused answer to AP Precalculus Topic 4.7, covering how to find parametric equations for an implicitly defined curve, the trig parametrization of circles and ellipses, and how to verify a parametrization satisfies the original equation.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

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Quick answer

Parametrizing an implicit curve means finding $x(t)$ and $y(t)$ so that, as $t$ varies, the point $(x(t), y(t))$ traces exactly the curve described by the implicit equation. The standard tool for conics is the Pythagorean identity $\cos^2 t + \sin^2 t = 1$ . A circle $x^2 + y^2 = r^2$ is parametrized by $x = r\cos t$ , $y = r\sin t$ . An ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is parametrized by $x = a\cos t$ , $y = b\sin t$ , using the semi-axes (the square roots of the denominators). A line or other curve can sometimes be parametrized by setting $x = t$ and solving for $y$ . You verify any parametrization by substituting $x(t)$ and $y(t)$ into the implicit equation and checking it reduces to a true statement for all $t$ . The parametrization adds direction and timing to a curve that the implicit equation alone does not specify.

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What this topic is asking
The idea of parametrizing a curve
Parametrizing circles and ellipses
Parametrizing other curves
Try this

What this topic is asking

The College Board (Topic 4.7) wants you to parametrize an implicitly defined curve: find parametric equations $x(t), y(t)$ whose points trace the curve given by an implicit equation. For circles and ellipses this uses the trigonometric parametrization and the Pythagorean identity, and you verify the result by substituting back into the original equation.

The idea of parametrizing a curve

The same curve can be parametrized many ways (different directions, speeds, starting points); any parametrization that satisfies the equation and traces the curve is valid.

Parametrizing circles and ellipses

Parametrizing and verifying an ellipse

Parametrize $\frac{(x - 1)^2}{16} + \frac{(y + 2)^2}{4} = 1$ and verify it.

step 1 Identify the center and semi-axes

Center $(1, -2)$ ; semi-axes $a = \sqrt{16} = 4$ horizontally and $b = \sqrt{4} = 2$ vertically.

step 2 Write the parametrization

$x(t) = 1 + 4\cos t$ , $y(t) = -2 + 2\sin t$ , for $0 \le t < 2\pi$ .

step 3 Substitute back

$\frac{(4\cos t)^2}{16} + \frac{(2\sin t)^2}{4} = \frac{16\cos^2 t}{16} + \frac{4\sin^2 t}{4} = \cos^2 t + \sin^2 t = 1$ . The equation holds for every $t$ .

step 4 Check a point

At $t = 0$ : $(1 + 4, -2 + 0) = (5, -2)$ , the rightmost point of the ellipse, $4$ units right of the center. The parametrization is correct.

Parametrizing other curves

Not every curve is a conic. A general approach is to set $x = t$ and solve the implicit equation for $y$ , giving $y(t)$ directly; this works whenever the equation can be solved for $y$ as a function. For curves that fail the vertical line test, you may need a parameter that is not $x$ itself, as with the trig parametrization of a circle, where the angle $t$ does the job. The choice of parameter is yours, as long as the resulting points satisfy the equation.

A point worth stating once is the role of verification. Because the same implicit equation has many possible parametrizations, the only sure check is substitution: plug $x(t)$ and $y(t)$ into the original equation and confirm it collapses to a true statement for all $t$ (for conics, via $\cos^2 t + \sin^2 t = 1$ ). A parametrization that "looks right" but fails this check traces a different curve. Making substitution a habit, rather than trusting the pattern, catches the common error of using the denominators instead of their square roots.

Try this

Q1. Parametrize the circle $x^2 + y^2 = 4$ . [1 point]

Cue. Radius $\sqrt{4} = 2$ : $x = 2\cos t$ , $y = 2\sin t$ .

Q2. What are the semi-axis coefficients for parametrizing $\frac{x^2}{25} + \frac{y^2}{9} = 1$ ? [1 point]

Cue. $\sqrt{25} = 5$ and $\sqrt{9} = 3$ , so $x = 5\cos t$ , $y = 3\sin t$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2025 (style)1 marksSection I, Part A (multiple choice, no calculator). Which parametrization traces the ellipse

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

? (A)

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

(B)

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

(C)

x = 3t,\ y = 2t

(D)

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

Show worked answer →

The correct answer is (B), $x = 3\cos t,\ y = 2\sin t$ .

Substitute into the ellipse: $\frac{(3\cos t)^2}{9} + \frac{(2\sin t)^2}{4} = \frac{9\cos^2 t}{9} + \frac{4\sin^2 t}{4} = \cos^2 t + \sin^2 t = 1$ . The semi-axes $3$ and $2$ are the square roots of the denominators; choice (A) uses the denominators themselves, which is wrong.

AP 2025 (style)4 marksSection II (free response, no calculator). The curve

x^2 + y^2 = 36

is given implicitly. (a) Write a parametrization

x(t), y(t)

for it. (b) Verify your parametrization satisfies the equation, and state where it starts at

t = 0

Show worked answer →

A 4-point question on parametrizing and verifying.

(a) Parametrization (2 points): the circle has radius $\sqrt{36} = 6$ , so $x(t) = 6\cos t$ , $y(t) = 6\sin t$ for $0 \le t < 2\pi$ .
(b) Verify and start (2 points): substitute: $(6\cos t)^2 + (6\sin t)^2 = 36\cos^2 t + 36\sin^2 t = 36(\cos^2 t + \sin^2 t) = 36$ , satisfying the equation. At $t = 0$ : $(6\cos 0, 6\sin 0) = (6, 0)$ , the rightmost point.

Related dot points

Sources & how we know this

AP Precalculus Course and Exam Description — College Board (2023)