What is -(8y+1)/(2x^2)=-(4y)/(x^2)-1/y ?

The solution to -(8y+1)/(2x^2)=-(4y)/(x^2)-1/y is Parabola with (h,k)=(0,0),p= 1/8

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-(8y+1)/(2x^2)=-(4y)/(x^2)-1/y

- \frac{8 y + 1}{2 x ^{2}} = - \frac{4 y}{x ^{2}} - \frac{1}{y}

(h, k) = (0, 0), p = \frac{1}{8}

Solution steps

- \frac{8 y + 1}{2 x ^{2}} = - \frac{4 y}{x ^{2}} - \frac{1}{y}

Rewrite

- \frac{8 y + 1}{2 x ^{2}} + \frac{4 y}{x ^{2}} + \frac{1}{y} = 0

in the standard form

4 \cdot \frac{1}{8} (y - 0) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = \frac{1}{8}

yy = \frac{8 x}{3} + {2}

y - 7 x^{2} = 2

4^{2} + 4 y^{2} - 20 x + 40 y = - 5

- x^{2} + 16 x + 32 y = 128

y^{2} - 4 x + 8 = 0

What is -(8y+1)/(2x^2)=-(4y)/(x^2)-1/y ?
The solution to -(8y+1)/(2x^2)=-(4y)/(x^2)-1/y is Parabola with (h,k)=(0,0),p= 1/8