Recipe

Analyze a conic from general form

Use when: you're handed a second-degree equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ and asked to identify the conic and pin down where it sits

Goal: the conic's family, plus its center, vertices, foci, or asymptotes read straight off a standard-form equation

Taught in Identifying a conic from its equation · a later section extends this recipe with more cases

Quick recipe

Collect every term onto one side
Check whether $B=0$
Apply the four-way test to $A$ and $C$
Complete the square in $x$ and in $y$
Read the center, vertices, foci, or asymptotes off the standard form

1
Move every term to one side so the equation reads $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ , flipping the sign of anything that crosses the equals sign.

Example

$3x^2+5=2y^2-6x+4y$ collects to $3x^2-2y^2+6x-4y+5=0$ .

Why: reading $A$ , $B$ , $C$ off an equation that still has terms split across both sides measures the wrong numbers entirely. The four-way test in Step 3 needs the real coefficients, not some tangled mix of them.
2
Look at the $xy$ -term's coefficient, $B$ .
- $B=0$ — the axes already line up with the conic; go straight to Step 3.
- $B\neq0$ — the conic is rotated. Either use the discriminant $B^2-4AC$ to name the family without rotating, or find the rotation angle from $\cot2\theta=\frac{A-C}{B}$ and substitute to eliminate the cross term before continuing.
- Not yet: $B\neq0$ — covered in Rotation of axes
Why: the four-way test in Step 3 reads $A$ and $C$ as if they're the only things shaping the curve. An uneliminated $xy$ term means $A$ and $C$ alone no longer tell the whole story.
3
Check in order: is exactly one of $A$ , $C$ zero? Parabola. Otherwise, opposite signs? Hyperbola. Otherwise, equal? Circle. Otherwise (same sign, unequal)? Ellipse.

Example

$9x^2+4y^2-18x+16y-11=0$ : $A=9$ , $C=4$ , same sign, unequal, so an ellipse.

Why: the sign pattern and equality of $A$ and $C$ alone fix the family, before any algebra runs -- the same test proved above.
4
Group the $x$ -terms and the $y$ -terms separately, factor $A$ and $C$ back out of their groups if either isn't $1$ , and complete the square inside each group.
- Not yet: the leading coefficients aren't both $1$ — covered in Translated conics and completing the square
Why: the general equation is a standard-form conic that's been multiplied out and shoved off-center; completing the square is exactly the un-scrambling move that reverses that expansion.
5
Match the completed-square equation to the family's standard form from earlier in this chapter and read off $(h,k)$ and whatever family-specific quantities the problem asks for.

Example

$x^2+y^2-4x+2y-4=0$ completes to $(x-2)^2+(y+1)^2=9$ : center $(2,-1)$ , radius $3$ .

Why: standard form is a spec sheet. Once an equation is in that shape, every quantity is a direct algebraic readout, not something left to solve for.

Common mistakes

Reading $A$ , $B$ , $C$ off an equation before collecting every term onto one side.
Running the four-way $A/C$ test on an equation that still has $B\neq0$ .
Forgetting that whatever's added inside a completed-square group is multiplied by the factored-out leading coefficient before it lands on the other side.
Reading a center's coordinate straight off $(x-h)^2$ without flipping the sign.
Treating the family test as proof a real curve exists: a same-sign, equal-coefficient pattern can still collapse to a single point or no graph at all, depending on the constant left over.