Conic Sections in the Cartesian Plane | Tony's Intertwingled Memex

Core Definition

A conic section is a plane curve obtained by intersecting a plane with a double-napped right circular cone. In Cartesian coordinates, all non-degenerate conic sections are algebraic curves of degree two described by the general bivariate quadratic equation $A x^2 + B xy + C y^2 + D x + E y + F = 0$ . Depending on the coefficients and eccentricity, they classify into circles, ellipses, parabolas, or hyperbolas.

Concept Explanation

In Cartesian coordinates without cross-terms ( $B = 0$ ), completing the square on quadratic expressions converts general equations into canonical standard forms that immediately reveal the curve's center/vertex, axes of symmetry, orientation, and asymptotic boundaries.

1. Circles

Standard Form: $(x - h)^2 + (y - k)^2 = r^2$
Key Features: Center $(h, k)$ , radius $r > 0$ . Both $x^2$ and $y^2$ have identical positive coefficients.

2. Ellipses

Horizontal Major Axis: $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \quad (a > b > 0)$ Center $(h, k)$ , major axis length $2a$ (horizontal), minor axis length $2b$ (vertical).
Vertical Major Axis: $\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1 \quad (a > b > 0)$ Center $(h, k)$ , major axis length $2a$ (vertical), minor axis length $2b$ (horizontal).
Identifying Clue: Both $x^2$ and $y^2$ terms are positive, but their denominators (coefficients) differ.

3. Parabolas

Vertical Axis (Opens Up/Down): $(x - h)^2 = 4p(y - k) \iff y - k = \frac{1}{4p}(x - h)^2$ Opens upward if $p > 0$ , downward if $p < 0$ . Vertex $(h, k)$ .
Horizontal Axis (Opens Left/Right): $(y - k)^2 = 4p(x - h) \iff x - h = \frac{1}{4p}(y - k)^2$ Opens rightward if $p > 0$ , leftward if $p < 0$ . Vertex $(h, k)$ .
Identifying Clue: Exactly one variable is squared, while the other variable appears only linearly to the first power.

4. Hyperbolas

Horizontal Transverse Axis (Opens Left & Right): $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$ Center $(h, k)$ , vertices $(h \pm a, k)$ , asymptotes $y - k = \pm \frac{b}{a}(x - h)$ .
Vertical Transverse Axis (Opens Up & Down): $\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$ Center $(h, k)$ , vertices $(h, k \pm a)$ , asymptotes $y - k = \pm \frac{a}{b}(x - h)$ .
Identifying Clue: The squared terms have opposite algebraic signs ( $+x^2$ and $-y^2$ , or $+y^2$ and $-x^2$ ).

When to Use It / Method Comparison

Feature / Clue	Circle	Ellipse	Parabola	Hyperbola
Squared Terms	Both $x^2$ and $y^2$	Both $x^2$ and $y^2$	Only one variable squared ( $x^2$ or $y^2$ )	Both $x^2$ and $y^2$
Coefficients of Squared Terms	Identical sign and magnitude ( $A = C > 0$ )	Identical sign, unequal magnitude ( $A \ne C$ , $AC > 0$ )	One squared coefficient is zero	Opposite algebraic signs ( $AC < 0$ )
Level Curve Prototype	$x^2 + y^2 = c$ (circular paraboloid slice)	$2x^2 + 3y^2 = c$ (elliptic cone/cylinder)	$z = y - x^2 \implies y = x^2 + c$ (parabolic gutter)	$x^2 - y^2 = c$ (hyperbolic paraboloid saddle)

Why We Use It in Calculus (Calculus III Context)

Multivariable Level Curves and Contour Maps: Setting $z = f(x, y) = c$ produces a family of 2D Cartesian curves. Recognizing standard conic forms allows instant sketching of topographical contours (e.g., concentric ellipses for paraboloid bowls or families of orthogonal hyperbolas for saddle points).
Domain Boundary Curves: When finding the domain of $f(x, y) = \sqrt{36 - 9x^2 - 4y^2}$ or $g(x, y) = \ln(y - x^2)$ , the boundary curve between defined and undefined regions is a conic section ( $\frac{x^2}{4} + \frac{y^2}{9} = 1$ or $y = x^2$ ).
Cross-Sectional Traces of Quadric Surfaces: Slicing 3D quadric surfaces (ellipsoids, hyperboloids of one/two sheets, hyperbolic paraboloids, cones) with planes $x = k$ , $y = k$ , or $z = k$ yields 2D conic section traces.

Engineering Applications

Aeronautical & Space Engineering: Orbital mechanics and spacecraft trajectory transfers utilize Keplerian orbits, which are conic sections with the celestial primary at one focus (ellipses for closed orbits, hyperbolas for planetary escape trajectories, parabolas for minimum-energy insertion).
Civil & Structural Engineering: Parabolic cable geometry in suspension bridges and parabolic arches equalize axial compression under uniformly distributed gravity loads.
Acoustics & Optical Engineering: Whispering galleries and reflecting telescopes leverage the reflective properties of elliptical and parabolic reflectors to focus sound and electromagnetic waves at precise focal points.

Example Problems

Example Problem 1: Converting General Form to Canonical Ellipse

Convert the following equation into canonical standard form, classify the conic section, and determine its center, vertices, and lengths of major and minor axes:

9x^2 + 25y^2 - 36x + 50y - 164 = 0

Solution

Step 1: Group variables and move constant to right-hand side

(9x^2 - 36x) + (25y^2 + 50y) = 164

Step 2: Factor out the leading coefficients from each group

9(x^2 - 4x) + 25(y^2 + 2y) = 164

Step 3: Complete the square inside each set of parentheses

For $x$ : half of $-4$ is $-2$ , $(-2)^2 = 4$ . Adding $4$ inside adds $9 \times 4 = 36$ to the equation.
For $y$ : half of $2$ is $1$ , $1^2 = 1$ . Adding $1$ inside adds $25 \times 1 = 25$ to the equation.

9(x^2 - 4x + 4) + 25(y^2 + 2y + 1) = 164 + 36 + 25

9(x - 2)^2 + 25(y + 1)^2 = 225

Step 4: Divide both sides by the constant term ( $225$ )

\frac{9(x - 2)^2}{225} + \frac{25(y + 1)^2}{225} = 1

\frac{(x - 2)^2}{25} + \frac{(y + 1)^2}{9} = 1

Step 5: Identify characteristics

Classification: Horizontal Ellipse ( $a^2 = 25 > b^2 = 9$ ).
Center: $(h, k) = (2, -1)$ .
Semi-major axis: $a = \sqrt{25} = 5 \implies$ Major axis length $2a = 10$ along the horizontal line $y = -1$ .
Semi-minor axis: $b = \sqrt{9} = 3 \implies$ Minor axis length $2b = 6$ along the vertical line $x = 2$ .
Vertices: $(h \pm a, k) = (2 \pm 5, -1) \implies (-3, -1)$ and $(7, -1)$ .

Example Problem 2: Multivariable Level Curve Identification

Given the two-variable saddle surface $z = f(x, y) = 4x^2 - y^2$ :

Determine the equation of the level curve at height $z = 16$ , classify the conic section, and state the equations of its asymptotes.
Determine the equation of the level curve at height $z = -4$ , classify the conic section, and state the direction in which its branches open.
Describe the trace in the $xy$ -plane at height $z = 0$ .

Solution

Step 1: Analyze level curve at $z = 16$ Set $f(x, y) = 16$ :

4x^2 - y^2 = 16

Divide through by

16

\frac{4x^2}{16} - \frac{y^2}{16} = 1 \implies \frac{x^2}{4} - \frac{y^2}{16} = 1

Classification: Hyperbola with horizontal transverse axis, centered at $(0, 0)$ .
Vertices: $(\pm 2, 0)$ , opening left and right.
Asymptotes: Setting the right-hand side to $0$ gives $\frac{x^2}{4} - \frac{y^2}{16} = 0 \implies y^2 = 4x^2 \implies y = \pm 2x$ .

Step 2: Analyze level curve at $z = -4$ Set $f(x, y) = -4$ :

4x^2 - y^2 = -4

Divide through by

-4

-\frac{4x^2}{4} + \frac{y^2}{4} = 1 \implies \frac{y^2}{4} - x^2 = 1

Classification: Hyperbola with vertical transverse axis, centered at $(0, 0)$ .
Vertices: $(0, \pm 2)$ , opening upward and downward.
Asymptotes: $y = \pm 2x$ .

Step 3: Analyze level curve at $z = 0$ Set $f(x, y) = 0$ :

4x^2 - y^2 = 0 \implies (2x - y)(2x + y) = 0

Description: A degenerate conic consisting of a pair of intersecting lines: $y = 2x$ and $y = -2x$ crossing at the origin $(0, 0)$ . This represents the asymptotic boundary dividing the horizontal and vertical hyperbolic level curves of the saddle surface.

Connections & References

Parent Concepts: Calculus Foundations
Prerequisite Technique: Completing the Square
Downstream Calculus Applications: