Posted by Anal-G on Friday, January 14, 2011 at 8:11am.
Your description defines an ellipse.
does the question expect you to find that equation by using that definition?
if so, then let such a point on that locus be P(x,y)
√[(x-2)^2 + y^2] + √[(x+2)^2 + y^2] = 8
√[(x-2)^2 + y^2] = 8 - √[(x+2)^2 + y^2]
square both sides and expand
x^2 - 4x + 4 + y^2 = 64 - 16√[(x+2)^2 + y^2] + x^2 + 4x + 4 + y^2
16√[(x+2)^2 + y^2] = 64 + 8x
2√[(x+2)^2 + y^2] = 8 + x
squaring again and simplifying I get
3x^2 + 4y^2 = 48
divide each term by 48 to get it into standard form
x^2/16 + y^2/12 = 1
or, the easy way
from the description 2a = 8 , a = 4
(2,0) and (-2,0) must be the focal points so the midpoint or (0,0) must be the centre and c = 2
since the focal points lie on the x-axis
b^2 + c^2 = a^2
b^2 + 4 = 16
b^2 = 12
standard form with centre (0,0) is
x^2/a^2 + y^2/b^2 = 1
so
x^2/16 +y^2/12 = 1
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