In AP Precalculus, an ellipse is a conic section centered at (h, k) with horizontal radius a and vertical radius b, written analytically as (x - h)²/a² + (y - k)²/b² = 1, and parametrized as x(t) = h + a cos t, y(t) = k + b sin t for 0 ≤ t ≤ 2π.
An ellipse is a stretched circle. The AP Precalculus CED defines it analytically as (x - h)²/a² + (y - k)²/b² = 1, where (h, k) is the center, a is the horizontal radius, and b is the vertical radius (EK 4.6.A.2). The numbers under each fraction control how far the curve reaches from the center. From (h, k), the ellipse extends a units left and right and b units up and down. A circle is just the special case where a = b, so if you can read a circle's equation, you already know how to read an ellipse's.
The ellipse also shows up in parametric form. Per EK 4.7.B.2, you can trace it with x(t) = h + a cos t and y(t) = k + b sin t for 0 ≤ t ≤ 2π. This works because of the Pythagorean identity. Plug the parametric equations into the standard form and you get cos²t + sin²t, which equals 1 for every value of t. That's the whole reason the parametrization satisfies the equation, and it's exactly the kind of reasoning AP Precalc wants you to be able to explain.
The ellipse lives in Unit 4 (Functions Involving Parameters, Vectors, and Matrices) and hits two learning objectives. Under AP Pre Calc 4.6.A, you represent conic sections with horizontal or vertical symmetry analytically, which means reading and writing the standard-form equation. Under AP Pre Calc 4.7.B, you represent conics parametrically, which for the ellipse means using cosine for x and sine for y. The ellipse is the bridge between these two skills. It's the cleanest example of an implicitly defined curve (no single function y = f(x) can describe the whole thing, since it fails the vertical line test) that becomes simple once you introduce a parameter t. That idea, trading one implicit equation for two explicit functions of t, is the core move of Topics 4.6-4.7.
Keep studying AP® Precalculus Unit 4
Hyperbola (Unit 4)
The hyperbola's equation is the ellipse's with one sign flipped, (x - h)²/a² minus (y - k)²/b² = 1 instead of plus. That single minus sign changes everything. The ellipse is a closed loop, while the hyperbola splits into two branches that run off toward asymptotes.
Parameter (Unit 4)
The ellipse is the showcase example of parametrization. The implicit equation tells you which points are on the curve, but the parametric form (h + a cos t, k + b sin t) tells you how to travel around it, starting at the rightmost point when t = 0 and going counterclockwise.
Parabola (Unit 4)
Both are conic sections in Topic 4.6, but a parabola can be solved for x or y, so it parametrizes the easy way as (t, f(t)). An ellipse can't be solved for y as a single function, which is exactly why it needs the trig parametrization instead.
Invertible function (Unit 4)
An ellipse fails the vertical line test, so it isn't the graph of a function at all, let alone an invertible one. Topic 4.7 uses curves like the ellipse to show why parametric representation exists. It handles curves that function notation can't.
Ellipse questions are mostly translation tasks in both directions. Given an equation like (x-3)²/25 + (y+4)²/9 = 1, you pull out the center (3, -4), the horizontal radius 5, and the vertical radius 3. Watch the two classic traps. The denominators are a² and b², not a and b, so 25 means radius 5. And the signs flip, so (y + 4) means k = -4. Going the other way, you build the equation from a described center and radii. Multiple-choice questions also like transformations, such as stretching a circle vertically by a factor of 2 and asking for the resulting ellipse's equation (only the b² value changes), or giving you a point on the ellipse plus a symmetry condition and asking you to find the center. For Topic 4.7, expect to convert between the standard form and the parametrization x(t) = h + a cos t, y(t) = k + b sin t, or to identify which curve a given parametrization traces.
The equations look nearly identical, but the ellipse adds its two fractions while the hyperbola subtracts. The plus sign in (x-h)²/a² + (y-k)²/b² = 1 gives you a closed, bounded oval. The minus sign in the hyperbola's equation gives you two open branches with asymptotes. On the exam, check the sign between the fractions first. It instantly tells you which conic you're looking at. Their parametrizations differ too. The ellipse uses cos t and sin t (which satisfy cos²t + sin²t = 1), while the hyperbola needs a different trig setup because its identity involves a difference.
The standard form of an ellipse is (x - h)²/a² + (y - k)²/b² = 1, where (h, k) is the center, a is the horizontal radius, and b is the vertical radius.
The denominators are the radii squared, so a denominator of 25 means a radius of 5, not 25.
A circle is a special case of an ellipse where a = b, so circles and ellipses share the same equation structure.
An ellipse can be parametrized as x(t) = h + a cos t and y(t) = k + b sin t for 0 ≤ t ≤ 2π, and substituting these into the standard form gives cos²t + sin²t = 1.
The sign inside each parenthesis is opposite the coordinate of the center, so (y + 4)² means k = -4.
A plus sign between the two fractions means ellipse; a minus sign means hyperbola.
It's a conic section centered at (h, k) with horizontal radius a and vertical radius b, written as (x - h)²/a² + (y - k)²/b² = 1. It appears in Topic 4.6 (analytic form) and Topic 4.7 (parametric form).
Yes. Per the CED (EK 4.6.A.2), a circle is the special case of an ellipse where a = b, meaning the horizontal and vertical radii are equal. Stretching a circle in one direction turns it into a non-circular ellipse.
The ellipse's equation adds the two fractions and produces a closed oval, while the hyperbola's equation subtracts them and produces two open branches with asymptotes. Check the sign between the fractions to tell them apart instantly.
Use x(t) = h + a cos t and y(t) = k + b sin t for 0 ≤ t ≤ 2π (EK 4.7.B.2). It works because substituting back into the standard form gives cos²t + sin²t, which equals 1 for every t.
No, and AP Precalc doesn't frame it that way. In the CED, a is simply the horizontal radius and b is the vertical radius. Either one can be larger, so don't assume a > b.
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