Parametric functions of conic sections — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Parametric equations of circles, ellipses, hyperbolas, and parabolas; converting between parametric and Cartesian forms; eliminating parameters; orientation of parametric conics; and solving intersection problems.

You should already know: Cartesian equations of all conic sections. Basic Pythagorean and trigonometric identities. Parametric function definition and notation.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Parametric functions of conic sections?

A parametric function of a conic section expresses the $x$- and $y$-coordinates of any point on the conic as separate functions of a common independent variable called the parameter (most often denoted $t$ or $\theta$). Unlike Cartesian form, which relates $x$ and $y$ directly, parametric form separates the two coordinates, making it easy to track position, orientation, and motion along the conic. According to the AP Precalculus Course and Exam Description (CED), this topic is part of Unit 4: Functions Involving Parameters, Vectors, and Matrices, which accounts for 20–30% of the total AP exam weight. Parametric conics typically appear in both multiple-choice (MCQ) and free-response (FRQ) sections, often paired with motion or intersection problems. Synonyms for this topic include parametric equations of conic sections and parametric conics. It is especially useful for real-world applications, as it lets you find an object’s position at a specific time directly without solving for one variable in terms of the other.

2. Standard Parametric Forms of Conic Sections

Each standard conic centered at $(h,k)$ has a widely used parametric form derived from Pythagorean trigonometric identities for closed conics (circles, ellipses) and hyperbolas, and from polynomial mapping for parabolas.

• For a circle with radius $r$ centered at $(h,k)$: $$x=h+r\cos t,y=k+r\sin t,0\le t<2\pi$$ The parameter $t$ corresponds to the angle from the positive $x$-axis, so increasing $t$ produces counterclockwise orientation starting at $(h+r,k)$.
• For an ellipse with horizontal major axis centered at $(h,k)$, semi-major axis $a$, semi-minor axis $b$: $$x=h+a\cos t,y=k+b\sin t,0\le t<2\pi$$ For a vertical major axis ellipse, swap $a$ and $b$: $a$ (the semi-major axis) goes in the $y$ equation instead of $x$.
• For a hyperbola with horizontal transverse axis centered at $(h,k)$, semi-transverse axis $a$, semi-conjugate axis $b$: $$x=h+a\sec t,y=k+b\tan t$$ For a vertical transverse axis hyperbola, swap roles: $x=h+b\tan t,y=k+a\sec t$, using the identity ${\sec }^{2}t-{\tan }^{2}t=1$.
• For an upward-opening parabola with vertex $(h,k)$ and focal length $p$: $x=h+2pt,y=k+p{t}^2$, a polynomial parameterization that simplifies eliminating the parameter.

Worked Example

Write the parametric equations for an ellipse centered at $(2,-3)$, with a vertical major axis of total length 10, minor axis of total length 6, oriented counterclockwise. Find the coordinates of the point when $t=\frac{\pi }{6}$.

1. Extract parameters from the problem: center $(h,k)=(2,-3)$. Total major axis length $2a=10⟹a=5$, total minor axis length $2b=6⟹b=3$.
2. For a vertical major axis ellipse, $a$ belongs to the $y$ equation, so we use the form $x=h+b\cos t$, $y=k+a\sin t$.
3. Substitute values to get the full parametric equation: $x(t)=2+3\cos t$, $y(t)=-3+5\sin t$, $0\le t<2\pi$.
4. Evaluate at $t=\frac{\pi }{6}$: $\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$, $\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$.
5. Calculate coordinates: $x=2+3\left(\frac{\sqrt{3}}{2}\right)=\frac{4+3\sqrt{3}}{2}\approx 4.60$, $y=-3+5\left(\frac{1}{2}\right)=-0.5$. The point is $\left(\frac{4+3\sqrt{3}}{2},-\frac{1}{2}\right)$.

Exam tip: Always confirm the orientation of the major/transverse axis before writing parametric equations. If the axis is vertical, swap $a$ to the $y$ equation—swapping incorrectly is the most common MCQ mistake on this skill.

3. Converting Between Parametric and Cartesian Forms (Eliminating the Parameter)

A core AP Precalculus skill is eliminating the parameter from a parametric conic to get its Cartesian equation, or writing a parametric equation from a given Cartesian conic. The process depends on the type of parameterization:

1. For trigonometric parameterizations (circles, ellipses, hyperbolas): Isolate the trigonometric term in each equation, then apply the relevant Pythagorean identity to eliminate $t$.
2. For polynomial parameterizations (most parabolas): Solve for $t$ from the linear (first-order) equation, then substitute into the quadratic equation to eliminate $t$.

This process works because all standard parametric forms are constructed to satisfy the Cartesian standard form by design, so you will end up with the correct standard form after simplification.

Worked Example

Eliminate the parameter $t$ from the parametric equations $x=1+4\sec t$, $y=-2+3\tan t$, and identify the type of conic and its key features.

1. Isolate the trigonometric terms on one side of each equation: $x-1=4\sec t⟹\sec t=\frac{x-1}{4}$ $y+2=3\tan t⟹\tan t=\frac{y+2}{3}$
2. Apply the hyperbolic Pythagorean identity ${\sec }^{2}t-{\tan }^{2}t=1$.
3. Substitute the expressions for $\sec t$ and $\tan t$ into the identity: $${\left(\frac{x-1}{4}\right)}^2-{\left(\frac{y+2}{3}\right)}^2=1$$
4. This matches the standard Cartesian form of a hyperbola centered at $(1,-2)$ with a horizontal transverse axis, semi-transverse axis $a=4$, and semi-conjugate axis $b=3$.

Exam tip: When eliminating the parameter for hyperbolas, don't mix up the order of the terms: the term corresponding to $\sec t$ is always the positive leading term, and the $\tan t$ term is always the subtracted second term. Reversing them gives a hyperbola opening the wrong direction.

4. Orientation and Intersection of Parametric Conics

Orientation is the direction a point moves along the conic as the parameter increases. For standard trigonometric parameterizations of closed conics (circles, ellipses), increasing $t$ from $0$ to $2\pi$ gives counterclockwise orientation starting at the rightmost point of the conic. To reverse orientation, replace $t$ with $-t$, which flips the sign of $\sin t$ (and leaves $\cos t$ unchanged) resulting in clockwise motion.

For intersection problems, the parameters of two different parametric curves are always independent, even if they are both named $t$ by default. To find intersections, set the $x$-coordinates equal and $y$-coordinates equal, use different variable names for each parameter, solve for both parameters, then find the intersection point.

Worked Example

A circle is given by $x=3\cos t$, $y=3\sin t$, and a line has parametric form $x=1+2s$, $y=1+2s$, with independent parameters $t$ and $s$. Find all intersection points.

1. Set $x$ equal and $y$ equal: $3\cos t=1+2s$ and $3\sin t=1+2s$.
2. Equate the right-hand sides to eliminate $s$: $3\cos t=3\sin t⟹\tan t=1$. This gives solutions $t=\frac{\pi }{4}$ and $t=\frac{5\pi }{4}$ for $0\le t<2\pi$.
3. Calculate coordinates from the circle equation: For $t=\frac{\pi }{4}$: $x=3\left(\frac{\sqrt{2}}{2}\right)=\frac{3\sqrt{2}}{2}$, $y=\frac{3\sqrt{2}}{2}$ For $t=\frac{5\pi }{4}$: $x=3\left(-\frac{\sqrt{2}}{2}\right)=-\frac{3\sqrt{2}}{2}$, $y=-\frac{3\sqrt{2}}{2}$
4. Verify that both points satisfy the line equation by solving for $s$: valid real solutions exist for both points, so both are intersections.

Exam tip: Always rename the parameter for the second curve when solving for intersections. Using the same parameter name for both curves almost always leads to missing solutions or extraneous points.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Swapping $a$ and $b$ when writing parametric equations for a vertical major axis ellipse, keeping $a$ in the $x$ equation instead of moving it to $y$. Why: Students memorize the horizontal form and forget to adjust for vertical orientation, matching what they remember instead of reading the problem. Correct move: Always note the axis direction first, place $a$ (the semi-axis matching the axis direction) in the equation for that coordinate.
• Wrong move: When eliminating the parameter from a hyperbola, reversing the order of terms so the ${\tan }^{2}t$ term is positive and the ${\sec }^{2}t$ term is subtracted. Why: Confusion between rearrangements of the identity ${\sec }^{2}t-{\tan }^{2}t=1$ leads to sign error. Correct move: Always write the squared $\sec t$ term first, subtract the squared $\tan t$ term, and set equal to 1.
• Wrong move: Assuming the parameter $t$ is the same for two intersecting parametric curves, setting $t_1=t_2=t$ when setting up the system. Why: Most problems default to $t$ as the parameter for both curves, leading students to assume they are the same variable. Correct move: Rename the second curve's parameter $s$ or $\theta$ before solving.
• Wrong move: Forgetting to add the domain restriction $0\le t<2\pi$ for circles and ellipses when writing full parametric equations. Why: Students focus on the $x(t)$ and $y(t)$ equations and ignore the parameter domain, leading to an incomplete description. Correct move: Always add the domain for closed conics when asked for a full parametric description.
• Wrong move: When eliminating the parameter from a parametric parabola, solving for $t$ from the quadratic $y$ equation instead of the linear $x$ equation. Why: Students don't notice which equation is linear, leading to unnecessary work and extraneous solutions. Correct move: Always solve for $t$ from the first-order (linear) equation and substitute into the second-order (quadratic) equation.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Which of the following is the parametric equation for a hyperbola centered at $(-1,4)$ with a vertical transverse axis, semi-transverse axis $a=2$, semi-conjugate axis $b=3$? A) $x=-1+3\tan t,y=4+2\sec t$ B) $x=-1+2\sec t,y=4+3\tan t$ C) $x=-1+3\cos t,y=4+2\sin t$ D) $x=-1+3\sec t,y=4+2\tan t$

Worked Solution: First, eliminate option C, which is the parametric form of an ellipse, not a hyperbola. For a vertical transverse axis hyperbola, the semi-transverse axis $a=2$ pairs with $\sec t$ in the $y$ equation, and the semi-conjugate axis $b=3$ pairs with $\tan t$ in the $x$ equation. This eliminates B (which has $\sec t$ in $x$, for a horizontal hyperbola) and D (which uses $a=3$ for the transverse axis). Only option A matches all requirements. The correct answer is A.

Question 2 (Free Response)

Consider the parametric equations $x(t)=3+2\cos t$, $y(t)=-1+4\sin t$, for $0\le t<2\pi$. (a) Identify the type of conic, and find its center and vertices. (b) Eliminate the parameter $t$ and write the equation in standard Cartesian form. (c) State the orientation of the conic and the starting point at $t=0$.

Worked Solution: (a) This matches the parametric form of an ellipse. The center is $(h,k)=(3,-1)$. Since $4>2$, the major axis is vertical, with semi-major axis $a=4$, so vertices are at $(3,-1\pm 4)=(3,3)$ and $(3,-5)$. (b) Isolate the trigonometric terms: $\frac{x-3}{2}=\cos t$ and $\frac{y+1}{4}=\sin t$. Use the Pythagorean identity ${\cos }^{2}t+{\sin }^{2}t=1$ to get the standard form: $${\left(\frac{x-3}{2}\right)}^2+{\left(\frac{y+1}{4}\right)}^2=1$$ (c) For standard trigonometric parameterization, the orientation is counterclockwise as $t$ increases. At $t=0$, $\cos 0=1$ and $\sin 0=0$, so the starting point is $(3+2(1),-1+4(0))=(5,-1)$.

Question 3 (Application / Real-World Style)

A satellite follows an elliptical orbit around Earth, with Earth centered at $(0,0)$ on a coordinate plane measured in thousands of kilometers. The ellipse has a horizontal major axis, with perigee (closest point to Earth) at $(4,0)$ and apogee (farthest point from Earth) at $(-8,0)$. Write parametric equations for the satellite's position over a 24-hour orbit, and find its position at $t=6$ hours.

Worked Solution: The center of the ellipse is at the midpoint of perigee and apogee: $h=\frac{4+(-8)}{2}=-2$, $k=0$. The total major axis length is $4-(-8)=12$, so semi-major axis $a=6$. For an ellipse, $c$ (distance from ellipse center to focus/Earth) is $|0-(-2)|=2$, so $b^2=a^2-c^2=36-4=32⟹b=4\sqrt{2}$. The parametric equations for a 24-hour orbit are: $$x(t)=-2+6\cos \left(\frac{2\pi t}{24}\right)=-2+6\cos \left(\frac{\pi t}{12}\right),y(t)=4\sqrt{2}\sin \left(\frac{\pi t}{12}\right),0\le t<24$$ At $t=6$, $\frac{\pi (6)}{12}=\frac{\pi }{2}$, so $\cos \left(\frac{\pi }{2}\right)=0$, $\sin \left(\frac{\pi }{2}\right)=1$. The position is $x=-2$, $y=4\sqrt{2}\approx 5.66$. In context, after 6 hours of orbit, the satellite is 2000 km west and 5660 km north of Earth's center.

7. Quick Reference Cheatsheet

8. What's Next

This chapter gives you the foundation to work with parametric motion, the next core topic in Unit 4. Parametric conics are most commonly used to model projectile motion (a parabolic path) and orbital motion (an elliptical path), so the next step is applying parametric derivatives to find velocity and acceleration of objects moving along conic paths. Without mastering converting between parametric and Cartesian forms, and finding intersections of parametric curves, you will not be able to solve motion problems that require finding position, speed, or collision events. This topic also feeds into vector-valued functions, where the position vector of a moving object is just a parametric representation of its path.

Parametric derivatives and motion Vector-valued functions Cartesian conic sections Implicit differentiation