What is (x-2)^2+(y-0)^2=(x+3)^2 ?

The solution to (x-2)^2+(y-0)^2=(x+3)^2 is Parabola with (h,k)=(0,0),p= 5/2

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(x-2)^2+(y-0)^2=(x+3)^2

(x - 2)^{2} + (y - 0)^{2} = (x + 3)^{2}

(h, k) = (0, 0), p = \frac{5}{2}

Solution steps

(x - 2)^{2} + (y - 0)^{2} = (x + 3)^{2}

Rewrite

(x - 2)^{2} + (y - 0)^{2} = (x + 3)^{2}

in the standard form

4 \cdot \frac{5}{2} (x - 0) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = \frac{5}{2}

y^{2} + 16 x - 6 y - 7 = 0

\frac{f ( x )}{2 x} \leq \frac{x ^{2} + 1}{2 x}

(y + 4)^{2} = - 42 (x + 2)

(\frac{2 y ^{2} + y}{4}) = x

y < x^{2} - x + 4

What is (x-2)^2+(y-0)^2=(x+3)^2 ?
The solution to (x-2)^2+(y-0)^2=(x+3)^2 is Parabola with (h,k)=(0,0),p= 5/2