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.

Question

I tried solving this but I got stuck with all the square roots and simplifying.

Answer 1

Looks like the definition of an ellipse to me. Let's see if it is

PF1 = √( (x-0)^2 + (y-2)^2 ) = √(x^2 + (y-2)^2)
PF2 = √( x^2 + (y+2)^2 )

√(x^2 + (y-2)^2) + √( x^2 + (y+2)^2 ) = 6
√(x^2 + (y-2)^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

Answer 2

last line should have been:
9x^2 + 5y^2 = 45 ----> looks like hx^2+ky^2=m

so it is an ellipse

Answer 3

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 = 45

Answer 4

okay, that makes sense. thank you!