PrecalculusConic Sections

Identifying Conics from Equations

Complete the square to name and graph any conic.

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Circles, ellipses, parabolas and hyperbolas all have equations of the form $Ax^2 + Cy^2 + Dx + Ey + F = 0$ once you multiply everything out. Given such an equation, you can name the conic at a glance, then complete the square to find its center or vertex and match a standard form from parabolas, ellipses or hyperbolas.

Key idea

For $Ax^2 + Cy^2 + Dx + Ey + F = 0$ (a genuine conic, not a degenerate one): $A = C$: circle; $A$ and $C$ the same sign: ellipse; opposite signs: hyperbola; only one squared term: parabola. If there is an $xy$ term, use $B^2 - 4AC$: negative for an ellipse, zero for a parabola, positive for a hyperbola.

Why it works

Completing the square only shifts the curve; it never changes the coefficients of $x^2$ and $y^2$. So those coefficients already tell you which standard form you will reach: two squares added (ellipse or circle), two squares subtracted (hyperbola), or one square (parabola). The $xy$ term rotates the curve, and $B^2 - 4AC$ is the combination that rotation does not change.
Switch between the three conics and compare how each is defined by distances.

Worked examples

Worked example: An ellipse

Identify $4x^2 + 9y^2 - 16x + 18y - 11 = 0$ and find its center.
  1. $A = 4$ and $C = 9$ have the same sign and are different: an ellipse.
  2. Group and factor: $4(x^2 - 4x) + 9(y^2 + 2y) = 11$.
  3. Complete the squares, adding $4 \cdot 4 = 16$ and $9 \cdot 1 = 9$ to both sides: $4(x - 2)^2 + 9(y + 1)^2 = 36$.
  4. Divide by $36$: $\frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{4} = 1$.

Answer: Ellipse with center $(2, -1)$

Worked example: A hyperbola

Identify $x^2 - y^2 + 4x + 2y - 1 = 0$ and find its center.
  1. $A = 1$, $C = -1$: opposite signs, a hyperbola.
  2. $$(x^2 + 4x + 4) - (y^2 - 2y + 1) = 1 + 4 - 1.$$
  3. $(x + 2)^2 - (y - 1)^2 = 4$.

Answer: Hyperbola with center $(-2, 1)$, opening left and right

Worked example: A parabola

Identify $y^2 - 4y - 8x + 20 = 0$ and find its vertex.
  1. Only $y$ is squared: a parabola.
  2. $y^2 - 4y + 4 = 8x - 20 + 4$, so $(y - 2)^2 = 8x - 16 = 8(x - 2)$.

Answer: Parabola with vertex $(2, 2)$, opening right

Common mistake

Completing the square inside a factored group but adding the wrong amount to the other side. In $4(x^2 - 4x + 4)$ the $+4$ inside the parentheses is really $+16$, because it is multiplied by $4$. Add $16$ to the other side, not $4$.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.

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Sources and further reading

  • OpenStax Precalculus 2e, Section 10.1 The Ellipse
  • OpenStax Precalculus 2e, Section 10.2 The Hyperbola
  • OpenStax Precalculus 2e, Section 10.3 The Parabola
  • OpenStax Precalculus 2e, Section 10.4 Rotation of Axes
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