Hyperbolas

Conic Sections

Overview Standard Hyperbolas Shifted Hyperbolas Conic Applications General Equation

General Equation

The general equation of a conic section is given below, where \(A, B, ..., F\) are Real numbers.

\[Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\]

However, the \(Bxy\) term can result in conics that are skewed or oriented in directions other than horizontal or vertical which is beyond the scope of this course, so we will focus on conics equations where \(B = 0\). This simplifies the above equation to the following.

We refer to the equation below as the general conic equation because this one equation is able to express all four primary conic sections (parabolas, circles, ellipses, and hyperbolas) as well as all degenerate conic sections (line, pair of intersecting lines, or single point).

\[Ax^2 + Cy^2 + Dx + Ey + F = 0\]

For a given conic equation in general form that is a non-degenerate conic, and assuming that both \(A \ne 0\) and \(C \ne 0\), we can easily determine which primary conic section is represented by the given equation using the following.

  1. A parabola is formed if either \(A=0\) or \(C=0\). In other words, the equation only has one squared term.
  2. A circle is formed if \(A=C\).
  3. An ellipse is formed if \(A\) and \(C\) have the same sign and \(A \ne C\).
  4. A hyperbola is formed if \(A\) and \(C\) have opposite sign.

Once we identify which conic section is represented by a given general equation, we can rewrite the general equation in the corresponding standard equation form by completing the square.

Here are a few examples of conics given in general form and their cooresponding standard form.

General vs. Standard
Graph
General Equation
Standard Equation
Reasoning
\[y^2+4x+10y-15=0\]
\[(y+5)^2 = -4(x-10)\]

The equation represents a parabola because \(A=0\), meaning the \(x^2\) term is missing.

\[x^2+y^2-10x-10y-14=0\]
\[(x-5)^2 + (y-5)^2 = 64\]

The equation represents a circle because \(A=C\), since \(A=1\) and \(C=1\).

\[x^2+4y^2+10x-40y+61=0\]
\[\frac{(x+5)^2}{64} + \frac{(y-5)^2}{16} = 1\]

The equation represents an ellipse because \(A\) and \(C\) have the same signs but are not equal, since \(A=1\) and \(C=4\).

\[-9x^2+36y^2-90x+360y+351=0\]
\[\frac{(y+5)^2}{9} - \frac{(x+5)^2}{36} = 1\]

The equation represents a hyperbola because \(A\) and \(C\) have opposite signs, since \(A=-9\) and \(C=36\).