What is (x-3)^2+(y-2)^2=(x+3)^2-(y+2)^2 ?

The solution to (x-3)^2+(y-2)^2=(x+3)^2-(y+2)^2 is Parabola with (h,k)=(2/3 ,0),p= 3/2

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(x-3)^2+(y-2)^2=(x+3)^2-(y+2)^2

(x - 3)^{2} + (y - 2)^{2} = (x + 3)^{2} - (y + 2)^{2}

(h, k) = (\frac{2}{3}, 0), p = \frac{3}{2}

Solution steps

(x - 3)^{2} + (y - 2)^{2} = (x + 3)^{2} - (y + 2)^{2}

Rewrite

(x - 3)^{2} + (y - 2)^{2} = (x + 3)^{2} - (y + 2)^{2}

in the standard form

4 \cdot \frac{3}{2} (x - \frac{2}{3}) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (\frac{2}{3}, 0), p = \frac{3}{2}

9 y = x^{2}

d x = \frac{6 y}{x}

3 x^{2} + 9 x - 5 y - 2 = 0

y = 20 x - 2 x^{2} + 10

(y + 3)^{2} = - 12 (x - 2)

What is (x-3)^2+(y-2)^2=(x+3)^2-(y+2)^2 ?
The solution to (x-3)^2+(y-2)^2=(x+3)^2-(y+2)^2 is Parabola with (h,k)=(2/3 ,0),p= 3/2