Conic sections
The idea
Slice a double cone with a plane and the cut edge traces a circle, ellipse, parabola, or hyperbola — the conic sections. Algebraically they are graphs of second-degree equations in x and y, and each carries a defining distance property: an ellipse is the set of points whose distances to two foci have a constant sum, a hyperbola a constant difference, and a parabola equal distances to a focus and a line. These curves earn their keep in physics — orbits are ellipses, headlight reflectors are parabolas, and some navigation systems run on hyperbolas.
Most problems hand you a scrambled equation and ask for the geometry hidden inside. The decoding tool is completing the square in x and y separately, which produces a standard form exposing the center, vertices, and foci. Classify the type early from the squared terms: both squares with same-sign coefficients give an ellipse (a circle when they are equal), opposite signs give a hyperbola, and a single squared variable gives a parabola. The frequent slip while completing the square: a constant added inside parentheses gets multiplied by the coefficient outside, so adding 9 inside 9(x² − 6x) really adds 81 to the equation, not 9.
Worked example
Identify the conic 9x² + 4y² − 54x + 8y + 49 = 0 and find its center, vertices, and foci.
- Both x² and y² appear with positive but unequal coefficients (9 and 4), so the curve is an ellipse. Group by variable: 9(x² − 6x) + 4(y² + 2y) + 49 = 0.
- Complete each square, compensating for the outside coefficients: 9(x² − 6x + 9) − 81 + 4(y² + 2y + 1) − 4 + 49 = 0, since the 9 inside the first group is worth 9 × 9 = 81 and the 1 inside the second is worth 4.
- Combine constants (−81 − 4 + 49 = −36) and move them across: 9(x − 3)² + 4(y + 1)² = 36. Dividing by 36 gives the standard form (x − 3)²/4 + (y + 1)²/9 = 1.
- The larger denominator, 9, sits under the y-term, so the major axis is vertical: center (3, −1), a = 3, b = 2, and the vertices lie 3 above and below the center at (3, 2) and (3, −4).
- Locate the foci on the major axis with c² = a² − b² = 9 − 4 = 5: c = √5 ≈ 2.24, so the foci are (3, −1 + √5) and (3, −1 − √5), safely inside the vertices as foci must be.
Answer. An ellipse centered at (3, −1) with a vertical major axis: vertices (3, 2) and (3, −4), foci (3, −1 + √5) and (3, −1 − √5).
Check your understanding
- How can you classify a conic from the signs and presence of its squared terms before completing any squares?
- Why does adding a constant inside 9(x² − 6x) change the equation by nine times that constant, and how do you compensate correctly?
- What happens to an ellipse as its two foci slide together into a single point?
- Why do ellipses use c² = a² − b² while hyperbolas use c² = a² + b², and what does the sign difference say about where the foci sit?
Build the foundations first
Conic sections builds on these concepts. If any feel shaky, start there.