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

The solution to (x^2-y)/(y^2+2xy)=0 is Parabola with (h,k)=(0,0),p= 1/4

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(x^2-y)/(y^2+2xy)=0

\frac{x ^{2} - y}{y ^{2} + 2 x y} = 0

(h, k) = (0, 0), p = \frac{1}{4}

Solution steps

\frac{x ^{2} - y}{y ^{2} + 2 x y} = 0

Rewrite

\frac{x ^{2} - y}{y ^{2} + 2 x y} = 0

in the standard form

4 \cdot \frac{1}{4} (y - 0) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = \frac{1}{4}

(y + 5)^{2} = 6 (x + 1)

(x + 2)^{2} = 66.6666666 (y - 0.94)

y^{2} - x - 49 = 0

y^{2} - 10 y - 12 x + 13 = 0

- 2 x^{2} - 2 y = 0

What is (x^2-y)/(y^2+2xy)=0 ?
The solution to (x^2-y)/(y^2+2xy)=0 is Parabola with (h,k)=(0,0),p= 1/4