EXAMPLE 1:
Find the standard equation of the ellipse in Figure 1 using the locus definition.
Figure 1
Since the horizontal axis of the ellipse is longer than the vertical axis of the ellipse, you use PF1 + PF2 = 2a.
Using the distance formula, you can find the length of PF1 and PF2.
  • Considering the points (-3, 0) and P(x, y), you can compute PF1 = [(x + 3)2 + (y - 0)2]½
  • Considering the points (3, 0) and P(x, y), you can compute PF2 = [(x - 3)2 + (y - 0)2]½
    • You also know that 2a = 10. Thus:
    PF1 + PF2 = 2a
    [(x + 3)2 + y2]½ + [(x - 3)2 + y2]½ = 10
    Isolate one of the radicals:
    [(x + 3)2 + y2]½ = 10 - [(x - 3)2 + y2]½
    Square both sides:
    (x + 3)2 + y2 = 100 - 20[(x - 3)2 + y2]½ + (x - 3)2 + y2
    Expand:
    x2 + 6x + 9 + y2 = 100 - 20[(x - 3)2 + y2]½ + x2 - 6x + 9 + y2
    Isolate the radical:
    20[(x - 3)2 + y2]½ = 100 -12x
    Square both sides:
    400(x - 3)2 + y2 = 10,000 - 2,400x + 144x2
    Expand:


    400(x2 - 6x + 9 + y2) = 10,000 - 2,400x + 144x2

    400x2 - 2400x + 3600 + 400y2 = 10,000 - 2,400x + 144x2
    Simplify:
    256x2 + 400y2 = 6400
    Write in standard form:
    x2/25 + y2/16 = 1
    Notice that x2/25 + y2/16 = 1 is an equation of the form x2/a2 + y2/b2 = 1.
    This is the standard equation of the above ellipse.