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.

asked by maath
  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

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

    so it is an ellipse

    posted by Reiny
  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

    posted by Steve
  4. okay, that makes sense. thank you!

    posted by maath

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