DWG NO. 8.4 — Lesson 4 of 4

Identifying and Classifying Conics from General Form

Unit 8: Conic Sections · ~25 min

Objective Classify a conic section from its general-form equation Ax² + Cy² + Dx + Ey + F = 0, and convert it to standard form by completing the square.

Every circle, ellipse, parabola, and hyperbola you've graphed this unit can be written, once multiplied out, as Ax² + Cy² + Dx + Ey + F = 0 — the same general form you used for circles back in Lesson 8.1, now covering the whole family. (This assumes the conic's axes line up with the x- and y-axes; a tilted conic would also need an xy-term, which this course doesn't cover.) The payoff is that you don't need to complete the square just to know what shape you're dealing with — the coefficients A and C alone give it away.

Reading A and C

A = C ≠ 0 — circle (equal coefficients mean equal "stretch" in both directions).
A ≠ C, same sign (both positive or both negative) — ellipse.
Exactly one of A, C is zero (the other nonzero) — parabola. Only one variable is squared.
A and C opposite signs (one positive, one negative, both nonzero) — hyperbola.

One more thing to watch for: completing the square can sometimes produce a degenerate case — a single point, a pair of intersecting lines, or no real graph at all (when a sum of squares is set equal to a negative number). The coefficient test above tells you the family; only finishing the algebra confirms you have an actual, non-degenerate curve.

4x² + 9y² − 16x + 54y + 61 = 0, converted below, graphs as the ellipse (x−2)\(\frac{^{2}}{9}\) + (y+3)\(\frac{^{2}}{4}\) = 1

Worked Example 1 · Classify and convert (ellipse)

ProblemClassify and convert 4x² + 9y² − 16x + 54y + 61 = 0.

1A = 4, C = 9 — same sign, different values, so this is an ellipse.

2Group and factor out leading coefficients: 4(x² − 4x) + 9(y² + 6y) = −61.

3Complete each square: 4(x²−4x+4) + 9(y²+6y+9) = −61 + 4(4) + 9(9) = −61+16+81 = 36.

44(x−2)² + 9(y+3)² = 36. Divide by 36: (x−2)\(\frac{^{2}}{9}\) + (y+3)\(\frac{^{2}}{4}\) = 1.

Ellipse, center (2, −3), a=3, b=2 (major axis horizontal)

Worked Example 2 · Classify and convert (hyperbola)

ProblemClassify and convert 2x² − 3y² + 8x + 6y − 1 = 0.

1A = 2, C = −3 — opposite signs, so this is a hyperbola.

2Group: 2(x² + 4x) − 3(y² − 2y) = 1.

3Complete each square: 2(x²+4x+4) − 3(y²−2y+1) = 1 + 2(4) − 3(1) = 1+8−3 = 6.

42(x+2)² − 3(y−1)² = 6. Divide by 6: (x+2)\(\frac{^{2}}{3}\) − (y−1)\(\frac{^{2}}{2}\) = 1.

Hyperbola, center (−2, 1), horizontal transverse axis

Guided practice

1.Classify (do not convert): x² + y² − 4x + 2y − 4 = 0.
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A = C = 1: circle.
2.Classify (do not convert): 9x² + 4y² − 36x + 8y + 4 = 0.
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A = 9, C = 4, same sign, different values: ellipse.
3.Classify (do not convert): x² − y² + 6x − 4y − 4 = 0.
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A = 1, C = −1, opposite signs: hyperbola.
4.Classify (do not convert): y² − 8x + 2y + 17 = 0.
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A = 0 (no x² term), C = 1: parabola.
5.Convert and classify x² + y² + 6x − 8y + 21 = 0. What do you find?
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A = C = 1, so the coefficient test predicts a circle. Completing the square: (x²+6x+9) + (y²−8y+16) = −21+9+16 → (x+3)²+(y−4)² = 4. That's a real circle: center (−3, 4), radius 2. (Had the right side worked out negative instead — a sum of two squares can never be negative — there would be no real graph at all, even though A = C would still say "circle." That's why finishing the algebra matters, not just reading the coefficients.)

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