math
Let F1 = (0,2), F2 = (0,2), and P = (x,y). Use the distance formula to convert the equation PF1 + PF2 = 6 into Cartesian form. Simplify your answer until it reaches the form hx^2+ky^2=m.
I tried solving this but I got stuck with all the square roots and simplifying.

Looks like the definition of an ellipse to me. Let's see if it is
PF1 = √( (x0)^2 + (y2)^2 ) = √(x^2 + (y2)^2)
PF2 = √( x^2 + (y+2)^2 )
√(x^2 + (y2)^2) + √( x^2 + (y+2)^2 ) = 6
√(x^2 + (y2)^2) = 6  √( x^2 + (y+2)^2 )
square both sides
x^2 + y^2  4y + 4 = 36  12√( x^2 + (y+2)^2 ) + x^2 + y^2 + 4y + 4
12√( x^2 + (y+2)^2 ) = 8y + 36
3√( x^2 + (y+2)^2 ) = 2y + 9
square again:
9(x^2 + (y+2)^2 ) = 4y^2 + 36y + 81
9x^2 + 9y^2 + 36y + 36 = 4y^2 + 36y + 81
9x^2  5y^2 = 45 > looks like hx^2+ky^2=m
ok, it was a hyperbola and not an ellipse since k < 0
posted by Reiny

last line should have been:
9x^2 + 5y^2 = 45 > looks like hx^2+ky^2=m
so it is an ellipseposted by Reiny

sorry. If PF1+PF2=6 it must be an ellipse.
Ahem...
9x^2 + 9y^2 + 36y + 36 = 4y^2 + 36y + 81
9x^2  5y^2 = 45
should be
9x^2 + 5y^2 = 45posted by Steve

okay, that makes sense. thank you!
posted by maath
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