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Reading Time: 5 min
Last Updated: September 7, 2026
Main Ideas: 5
Reading Time: 5 min
Last Updated: September 7, 2026
Main Ideas: 5

Topic 4.6 Notes – Conic Sections

Verified for 2027 AP® Precalculus Exam
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Conic sections are graphs made from quadratic relations in both xx and yy. In this topic, you work only with axis-aligned, nondegenerate conics, which means parabola, ellipse, circle, and hyperbola, all without rotation. The main job is to recognize the type, rewrite to standard form, and read or build the equation from graph features.

What Conic Sections Are

A conic is usually a relation, not a single function of xx. That matters because circles, ellipses, and many hyperbolas fail the vertical line test, even though they still have equations and graphs.

Here’s the recognition pattern after you simplify into standard form:

  • One squared variable only →\rightarrow parabola
  • Both squared, same sign →\rightarrow ellipse
    • Circle is the special ellipse where the horizontal and vertical radii match
  • Both squared, opposite signs →\rightarrow hyperbola

Translations work the same way as in other graphing topics:

  • (x−h)(x-h) and (y−k)(y-k) point to (h,k)(h,k)
  • Read inside signs oppositely, so x+3=x−(−3)x+3=x-(-3)

For ellipses, circles, and hyperbolas, (h,k)(h,k) is the center. For a parabola, it is the vertex.

A common test mistake is classifying from a scrambled equation too early. Rewrite first.

Standard Forms and Key Features

Keep this reference image in mind as you review the standard forms. It shows the basic axis-aligned conics centered at the origin, which matches the parent forms before any shifts.

Study guide illustration

Parabolas

(y−k)=a(x−h)2 (y-k)=a(x-h)^2

  • vertex (h,k)(h,k)
  • axis of symmetry x=hx=h
  • a>0a>0 opens up, a<0a<0 opens down

(x−h)=a(y−k)2 (x-h)=a(y-k)^2

  • vertex (h,k)(h,k)
  • axis of symmetry y=ky=k
  • a>0a>0 opens right, a<0a<0 opens left

Only the variable perpendicular to the opening direction is squared. Also, larger ∣a∣|a| means narrower.

Ellipses and circles

(x−h)2a2+(y−k)2b2=1 \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1

  • center (h,k)(h,k)
  • horizontal radius aa, vertical radius bb
  • endpoints (h±a,k)(h\pm a,k) and (h,k±b)(h,k\pm b)
  • symmetry lines x=hx=h and y=ky=k

(x−h)2+(y−k)2=r2 (x-h)^2+(y-k)^2=r^2

  • center (h,k)(h,k), radius rr
  • special case of ellipse with a=b=ra=b=r

Hyperbolas

(x−h)2a2−(y−k)2b2=1 \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1

  • center (h,k)(h,k), opens left and right
  • vertices (h±a,k)(h\pm a,k)

(y−k)2b2−(x−h)2a2=1 \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1

  • center (h,k)(h,k), opens up and down
  • vertices (h,k±b)(h,k\pm b)

For both forms,

y−k=±ba(x−h) y-k=\pm \frac{b}{a}(x-h)

Those are the asymptotes. In the reference image, that matches the hyperbola panel with the dashed diagonal lines. The opening direction comes from the positive squared term, not from the bigger denominator.

Rewriting to Standard Form

The algebra pattern is the same each time:

  1. Group xx-terms and yy-terms, move constants.
  2. Factor first if a leading coefficient is in the way.
  3. Complete the square using u2+Bu=(u+B2)2−(B2)2 u^2+Bu=\left(u+\frac{B}{2}\right)^2-\left(\frac{B}{2}\right)^2
  4. Simplify.
  5. For ellipses and hyperbolas, divide so the right side is 11.

Example parabola:

2x2−8x−y+11=0 2x^2-8x-y+11=0

y=2x2−8x+11=2(x2−4x)+11=2(x−2)2+3 y=2x^2-8x+11=2(x^2-4x)+11=2(x-2)^2+3

So

y−3=2(x−2)2 y-3=2(x-2)^2

It’s a parabola with vertex (2,3)(2,3), opening up.

Building an Equation from a Graph or Description

Write the correct standard form first, then fill in features.

  • Parabola with given vertex and point
    Vertex (1,−2)(1,-2), point (3,6)(3,6)
    y+2=a(x−1)2 y+2=a(x-1)^2
    8=4a⇒a=2 8=4a \Rightarrow a=2
    Equation is y+2=2(x−1)2y+2=2(x-1)^2
  • Ellipse with center and radii
    Center (−2,1)(-2,1), horizontal radius 55, vertical radius 33
    (x+2)225+(y−1)29=1 \frac{(x+2)^2}{25}+\frac{(y-1)^2}{9}=1
  • Circle with center and radius
    Center (4,−1)(4,-1), radius 66
    (x−4)2+(y+1)2=36 (x-4)^2+(y+1)^2=36
  • Hyperbola with center, vertices, asymptotes
    Center (3,−1)(3,-1), vertical vertices (3,−3)(3,-3) and (3,1)(3,1), slopes ±23\pm \frac23
    Here b=2b=2, and ba=23\frac{b}{a}=\frac23 so a=3a=3
    (y+1)24−(x−3)29=1 \frac{(y+1)^2}{4}-\frac{(x-3)^2}{9}=1

Graph Clues, Symmetry, and Common Mistakes

  • Ellipse sketch comes from center plus four endpoints.
  • Hyperbola sketch comes from the guiding rectangle x=h±ax=h\pm a, y=k±by=k\pm b, then the diagonals as asymptotes.

Hyperbola sketch from a guiding rectangle

  • Intercepts come from setting x=0x=0 or y=0y=0. Some conics have none.
  • Symmetry
    • parabola has one symmetry line
    • ellipse and hyperbola have symmetry across x=hx=h and y=ky=k
    • circle is symmetric across every line through its center

Common mistakes you want to avoid:

  • reading (x−h)(x-h) and (y−k)(y-k) with the wrong sign
  • assuming the larger denominator controls hyperbola opening
  • mixing up what aa means in a parabola versus an ellipse/hyperbola
  • calling every equation a function
  • classifying before rewriting to standard form

Key Takeaways

A conic should be classified from an equivalent standard form, not from the original unsimplified equation.
In a parabola, the squared variable tells you the axis of symmetry, and the graph opens in the other direction.
In a hyperbola, the positive squared term determines whether it opens horizontally or vertically.
For ellipses and hyperbolas, aa is the horizontal scale and bb is the vertical scale, even when the yy-term is written first.
The center or vertex comes from (x−h)(x-h) and (y−k)(y-k), with signs read oppositely.
Hyperbola asymptotes always use y−k=±ba(x−h)y-k=\pm \frac{b}{a}(x-h) in this topic.
A circle is just an ellipse with equal radii, so (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2 fits the same conic idea.

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