Two squared terms or one? Same sign or opposite? Same coefficient or not? Three questions, four answers.
Identify the conic
4x² − 9y² + 8x + 36y − 68 = 0
Both x² and y² are present, and their signs are opposite — positive 4 and negative 9. Opposite signs always mean a hyperbola. You never have to complete a single square to answer this; the two leading coefficients decide it on their own.
A missing square is the loudest signal in the whole topic. If only one variable is squared, stop: it is a parabola, and nothing else it can be.
A circle is the special ellipse where a = b, so it is no surprise that the test for it is the coefficients being equal. Divide 2x² + 2y² = 18 by 2 and it is plainly x² + y² = 9.
9x² + 4y² − 18x + 16y − 11 = 0
look only at 9x² and 4y²
both present, both positive
circle or ellipse
9 ≠ 4
so an ellipse
The −18x and +16y only move the centre. They can never turn an ellipse into a hyperbola, so they are simply not part of the identification.
x² − y² = 0 passes every hyperbola test and yet graphs as two straight lines. The signs tell you which family an equation belongs to; only completing the square tells you whether it has an honest curve.
The sign test only works on an equation with everything on one side. These three test that habit.
Squared terms on opposite sides: 3x² = 12 + 2y²
3x² − 2y² = 12
bring 2y² across; its sign flips
opposite signs
hyperbola
Read before moving and the "+" suggests an ellipse. Gather the squared terms on one side first.
Both squared terms negative: −x² − 4y² + 2x + 15 = 0
x² + 4y² − 2x − 15 = 0
multiply every term by −1
same sign, 1 ≠ 4
ellipse
Two negatives are still the SAME sign. Only a difference in sign makes a hyperbola.
Ellipse-shaped, but is it real? x² + 4y² − 2x + 8y + 5 = 0
(x² − 2x + 1) + 4(y² + 2y + 1) = −5 + 1 + 4
complete both squares; the 4 multiplies the second
(x − 1)² + 4(y + 1)² = 0
the right side is zero
a single point, (1, −1)
a degenerate ellipse
The type test said ellipse, and it is one, but with nothing inside. Completing the square is what showed it.
squared terms: y² only
no x² anywhere
so it is a parabola
question one settled it
(y − 1)² = −8x + 16 = −8(x − 2)
complete the square if details are wanted
vertex (2, 1), opens left
c = 2
Tests give a general equation and ask which conic it is, often with linear terms added to distract, and sometimes ask what x² − y² = 0 or a zero right-hand side really graphs as. The fastest route: put everything on one side, then look only at the x² and y² terms — count them, compare signs, compare coefficients. A 10-second check: a parabola has exactly one squared variable; if both are there, it cannot be one.
Identify the conic
5x² + 5y² − 20x + 30y + 45 = 0
Both squares are present with the same sign, and the coefficients are equal — 5 and 5 — so it is a circle. Dividing through by 5 gives x² + y² − 4x + 6y + 9 = 0, and completing the square gives centre (2, −3) with radius 2. A would need the coefficients to differ.
Hint: look only at 5x² and 5y². The linear terms and the constant cannot change which family the equation belongs to.
