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As with the ellipse, the ancient Greeks were familiar with the shape of hyperbolas and how to define them using the locus definition. Again, it was not until Descartes placed conic sections onto coordinate systems that equations for hyperbolas were developed and became meaningful.

Below you will see an example of how to arrive at the standard equation of hyperbolas starting with the locus definition and the special property |PF1 - PF2| = 2a.

EXAMPLE
Determine the standard equation of a hyperbola given that the focal points have coordinates (-3, 0) and (3, 0) and the vertices have coordinates (-2, 0) and (2, 0).
Use the fact we derived from the locus definition earlier, that |PF2 - PF1| = 2a, where a is the distance from the origin to one of the vertices.
Since |PF2 - PF1| = 2a, it is also true that PF2 - PF1 = ±2a.
Let's apply this formula to our example: Apply distance formula: [(x - 3)2 + y2]½ - [(x + 3)2 + y2]½= ±4 Isolate one radical: [(x - 3)2 + y2]½ = ±4 + [(x + 3)2 + y2]½ Square both sides: (x - 3)2 + y2 = 16 ± 8[(x + 3)2 + y2]½ + (x + 3)2 + y2 Expand: x2 - 6x + 9 = 16 ± 8[(x + 3)2 + y2]½ + x2 + 6x + 9 Simplify: -12x - 16 = ±8[(x + 3)2 + y2]½ 3x + 4 = ±2[(x + 3)2 + y2]½ Square both sides again: 9x2 + 24x + 16 = 4x2 + 24x + 36 + 4y2 Simplify: 5x2 - 4y2 = 20 Put in standard form: x2/ 4 - y2/ 5 = 1

Let's look now at how to use this same method to derive the standard equation for hyperbolas.