Identifying Conic Sections

A general second-degree equation takes the form: $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$

In most Grade 11 problems, the $xy$ term is missing ($B = 0$), leaving us with: $Ax^2 + Cy^2 + Dx + Ey + F = 0$

You can determine whether this equation represents a circle, ellipse, parabola, or hyperbola by examining the coefficients $A$ and $C$, or by completing the square to rewrite the equation in its standard form.

Using Coefficients to Identify Conics

When $B = 0$, you can quickly classify the conic section by looking at the squared terms:

• Parabola: Either $A = 0$ or $C = 0$ (only one squared variable).
• Circle: $A = C$ (both coefficients are exactly the same).
• Ellipse: $A$ and $C$ have the same sign but different values ($A \cdot C > 0$ and $A \neq C$).
• Hyperbola: $A$ and $C$ have opposite signs ($A \cdot C < 0$).

Completing the Square to Find Standard Form

While the quick test tells you the type of conic, completing the square gives you the standard form, revealing the center, vertices, and other key features.

1. Group the $x$ terms and $y$ terms together.
2. Factor out the leading coefficients $A$ and $C$ from the grouped terms.
3. Complete the square for both $x$ and $y$. Remember to add the equivalent value to the other side of the equation!
4. Divide by the constant term on the right side to set the equation to $1$ (for ellipses and hyperbolas).

Example 1: Identifying an Ellipse

Identify and convert to standard form: $4x^2 + 9y^2 - 16x + 18y - 11 = 0$

1. Quick Check: $A = 4$ and $C = 9$. Since they are both positive but not equal, this is an ellipse.

2. Complete the Square: Group terms: $(4x^2 - 16x) + (9y^2 + 18y) = 11$

Factor out coefficients: $4(x^2 - 4x) + 9(y^2 + 2y) = 11$

Complete the square (add $(-4/2)^2 = 4$ inside the first parenthesis, and $(2/2)^2 = 1$ inside the second). Add $4(4) = 16$ and $9(1) = 9$ to the right side: $4(x^2 - 4x + 4) + 9(y^2 + 2y + 1) = 11 + 16 + 9$ $4(x - 2)^2 + 9(y + 1)^2 = 36$

Divide by 36 to get 1 on the right side: $\frac{4(x - 2)^2}{36} + \frac{9(y + 1)^2}{36} = 1$ $\frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{4} = 1$

Example 2: Identifying a Hyperbola

Classify and convert to standard form: $x^2 - 4y^2 + 2x + 8y - 7 = 0$

1. Quick Check: $A = 1$ and $C = -4$. Since they have opposite signs, this is a hyperbola.

2. Complete the Square: Group terms: $(x^2 + 2x) - (4y^2 - 8y) = 7$ (Note: factoring out the negative changes the sign of the $y$ term!)

Factor out coefficients: $(x^2 + 2x) - 4(y^2 - 2y) = 7$

Complete the square. Add $1$ for $x$, and add $-4(1) = -4$ to the right side for $y$: $(x^2 + 2x + 1) - 4(y^2 - 2y + 1) = 7 + 1 - 4$ $(x + 1)^2 - 4(y - 1)^2 = 4$

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