What is x-y-2x+4y^2-2=0 ?

The solution to x-y-2x+4y^2-2=0 is Parabola with (h,k)=(-33/16 , 1/8),p= 1/16

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x-y-2x+4y^2-2=0

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

(h, k) = (- \frac{33}{16}, \frac{1}{8}), p = \frac{1}{16}

Solution steps

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

Rewrite

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

in the standard form

4 \cdot \frac{1}{16} (x - (- \frac{33}{16})) = (y - \frac{1}{8})^{2}

Therefore parabola properties are:

(h, k) = (- \frac{33}{16}, \frac{1}{8}), p = \frac{1}{16}

2 y^{2} + 8 y - x + 7 = 0

(y - 4)^{2} = (x - 3)^{2} + (y - 2)^{2}

x^{2} - y \leq 6

(y - 1)^{2} = 2 x

x^{2} - y \leq 2

What is x-y-2x+4y^2-2=0 ?
The solution to x-y-2x+4y^2-2=0 is Parabola with (h,k)=(-33/16 , 1/8),p= 1/16