Identifying and Graphing Conics from General Form

By the end of this lesson, you will be able to classify any conic from the general form $Ax^2 + Cy^2 + Dx + Ey + F = 0$ by completing the square to convert to standard form, and sketch the result.

The Big Idea

Real-world equations rarely arrive in standard form. The general form hides all the information — center, foci, vertices — until you complete the square for both $x$ and $y$ simultaneously. This lesson consolidates the entire conic section unit into one technique.

Classification (Review)

For $Ax^2 + Cy^2 + Dx + Ey + F = 0$ (with $B = 0$):

$A$ and $C$	Classification
$A = C \neq 0$	Circle
$A \neq C$, same sign	Ellipse
$A$ or $C = 0$ (not both)	Parabola
$A$ and $C$ opposite signs	Hyperbola
Both 0	Line (degenerate)

Complete the Square Strategy

1. Group $x$-terms and $y$-terms separately
2. Factor out coefficients before completing the square
3. Add the same values to both sides
4. Write in standard form and identify the conic

Example 1 — Circle

$x^2 + y^2 - 4x + 6y - 3 = 0$

$(x^2 - 4x + 4) + (y^2 + 6y + 9) = 3 + 4 + 9$

$(x-2)^2 + (y+3)^2 = 16$

Circle: center $(2,-3)$, radius $4$.

Example 2 — Ellipse

$9x^2 + 4y^2 - 36x + 8y + 4 = 0$

$9(x^2 - 4x) + 4(y^2 + 2y) = -4$

$9(x^2 - 4x + 4) + 4(y^2 + 2y + 1) = -4 + 36 + 4$

$9(x-2)^2 + 4(y+1)^2 = 36$

$$\frac{(x-2)^2}{4} + \frac{(y+1)^2}{9} = 1$$

Ellipse: center $(2,-1)$; $a^2 = 9$ under $y$ → vertical; $a = 3$, $b = 2$, $c = \sqrt{5}$.

Example 3 — Hyperbola

$4x^2 - y^2 + 32x + 6y + 39 = 0$

$4(x^2 + 8x) - (y^2 - 6y) = -39$

$4(x^2 + 8x + 16) - (y^2 - 6y + 9) = -39 + 64 - 9$

$4(x+4)^2 - (y-3)^2 = 16$

$$\frac{(x+4)^2}{4} - \frac{(y-3)^2}{16} = 1$$

Horizontal hyperbola: center $(-4,3)$; $a = 2$, $b = 4$, $c = \sqrt{20} = 2\sqrt{5}$.