A hyperbola is a conic section centered at (h, k) with two branches and two lines of symmetry, written as (x−h)²/a² − (y−k)²/b² = 1 if it opens left and right, or (y−k)²/b² − (x−h)²/a² = 1 if it opens up and down. The minus sign between the squared terms is what makes it a hyperbola instead of an ellipse.
A hyperbola is the conic section with two separate branches that open away from each other. In AP Precalculus (EK 4.6.A.3), you work with hyperbolas centered at (h, k) that have horizontal and vertical lines of symmetry. There are exactly two standard forms to know. If the x-term comes first, (x − h)²/a² − (y − k)²/b² = 1, the branches open left and right. If the y-term comes first, (y − k)²/b² − (x − h)²/a² = 1, the branches open up and down. Whichever variable is positive tells you the direction the hyperbola opens.
The feature that makes hyperbolas different from every other conic is their asymptotes. Both branches get closer and closer to two straight lines that cross at the center. For a left-right opening hyperbola, those asymptotes have slopes ±b/a and pass through (h, k). Think of the asymptotes as the skeleton of the graph. Once you sketch them, the branches just hug those lines. In Topic 4.7 (EK 4.7.B.3), you also see hyperbolas written parametrically using the trigonometric functions secant and tangent, the same way an ellipse is parametrized with cosine and sine.
The hyperbola lives in Unit 4 (Functions Involving Parameters, Vectors, and Matrices), specifically Topic 4.6 (Conic Sections) and Topic 4.7 (Parametrization of Implicitly Defined Functions). It directly supports learning objective AP Pre Calc 4.6.A, representing conic sections analytically, and AP Pre Calc 4.7.B, representing conic sections parametrically. The hyperbola is also your first serious example of an implicitly defined curve, meaning an equation in x and y that is not a function (it fails the vertical line test). That idea, handling curves that aren't y = f(x) by using two variables or a parameter t, is exactly the skill parametrization (LO 4.7.A) is built to solve, and it's the mindset you'll need for parametric and polar curves in calculus.
Keep studying AP® Precalculus Unit 4
Ellipse (Unit 4)
The ellipse and hyperbola equations are nearly identical. Swap the plus sign in (x − h)²/a² + (y − k)²/b² = 1 for a minus sign and you get a hyperbola. That one sign change turns a closed oval into two open branches, so checking the sign between the squared terms is the fastest conic ID move on a multiple-choice question.
Asymptote (Units 1 and 4)
You first met asymptotes with rational functions in Unit 1, where a graph approaches a line but never settles on it. A hyperbola brings that idea back with two slanted asymptotes crossing at the center, with slopes ±b/a for a left-right opener. Same concept, new shape.
Parameter (Unit 4)
A hyperbola is not a function, so you can't write it as one y = f(x). Parametrization fixes that by writing x and y each as functions of t. Per EK 4.7.B.3, a hyperbola is parametrized with secant and tangent, mirroring how an ellipse uses cosine and sine (EK 4.7.B.2). The identity sec²t − tan²t = 1 is doing the same job for hyperbolas that cos²t + sin²t = 1 does for ellipses.
Parabola (Unit 4)
The parabola is the other 'opens in a direction' conic in Topic 4.6, but it has only one squared variable, like x − h = a(y − k)². A hyperbola has two squared terms separated by a minus sign. One square means parabola, two squares with a minus means hyperbola.
Heads up on scope first. Unit 4 content, including conic sections, is not assessed on the AP Precalculus exam itself. The College Board includes it in the CED because it's essential preparation for calculus, and most teachers cover and test it in class. When hyperbolas show up on class assessments and practice questions, the tasks are predictable. You'll be asked to write the equation of a hyperbola from given features (center, a, b, foci, or the length of the transverse axis), to convert a general-form equation like 9x² − 4y² − 36x + 16y + 36 = 0 into standard form by completing the square so you can read off the center, and to find or use the asymptotes. A classic reverse problem gives you the asymptotes, say y = 2x + 3 and y = −2x + 3 with center (0, 3), and asks you to rebuild the equation from the slopes ±b/a. In Topic 4.7, expect to match a hyperbola to its parametrization using secant and tangent.
Both are conics centered at (h, k) with the same (x − h)²/a² and (y − k)²/b² building blocks, so they look like twins on paper. The difference is the sign between the squared terms. A plus sign gives an ellipse, a closed curve where a and b are the horizontal and vertical radii. A minus sign gives a hyperbola, two open branches with asymptotes. The sign also changes the parametrization, with cos t and sin t for the ellipse versus sec t and tan t for the hyperbola.
A hyperbola centered at (h, k) is written as (x − h)²/a² − (y − k)²/b² = 1 if it opens left and right, or (y − k)²/b² − (x − h)²/a² = 1 if it opens up and down.
The positive squared term tells you the opening direction, so a positive x-term means the branches open horizontally and a positive y-term means they open vertically.
The asymptotes pass through the center (h, k), and for a left-right opening hyperbola they have slopes ±b/a, which lets you rebuild the equation from the asymptotes alone.
To find the center of a hyperbola given in expanded general form, complete the square in both x and y to rewrite it in standard form.
A hyperbola is parametrized with secant and tangent (EK 4.7.B.3), just as an ellipse is parametrized with cosine and sine.
The minus sign between the squared terms is the entire difference between a hyperbola and an ellipse, so check the sign before anything else.
It's the conic section with two branches, defined in EK 4.6.A.3 as (x − h)²/a² − (y − k)²/b² = 1 (opens left-right) or (y − k)²/b² − (x − h)²/a² = 1 (opens up-down), centered at (h, k) with horizontal and vertical lines of symmetry.
No. Hyperbolas are in Unit 4 (Topics 4.6 and 4.7), and Unit 4 is not assessed on the AP Precalc exam. The College Board includes it as calculus preparation, so expect it on in-class tests and to need it next year.
One sign. An ellipse adds the two squared terms, (x − h)²/a² + (y − k)²/b² = 1, and forms a closed oval. A hyperbola subtracts them and forms two open branches with asymptotes. Their parametrizations differ the same way, cos/sin for ellipses versus sec/tan for hyperbolas.
Get the equation into standard form (complete the square if needed), find the center (h, k), then draw lines through the center with slopes ±b/a for a left-right opener. For example, 9x² − 4y² − 36x + 16y + 36 = 0 must be rewritten by completing the square before you can read off the center and slopes.
Use secant and tangent (EK 4.7.B.3), the hyperbola's version of the cos/sin parametrization of an ellipse. It works because sec²t − tan²t = 1 matches the minus sign in the hyperbola's equation, just as cos²t + sin²t = 1 matches the plus sign in an ellipse's.
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