What is 1/3 y^2-5/6 y+2x+1/6 =0 ?

The solution to 1/3 y^2-5/6 y+2x+1/6 =0 is Parabola with (h,k)=(17/96 , 5/4),p=-3/2

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1/3 y^2-5/6 y+2x+1/6 =0

\frac{1}{3} y^{2} - \frac{5}{6} y + 2 x + \frac{1}{6} = 0

(h, k) = (\frac{17}{96}, \frac{5}{4}), p = - \frac{3}{2}

Solution steps

\frac{1}{3} y^{2} - \frac{5}{6} y + 2 x + \frac{1}{6} = 0

Rewrite

\frac{1}{3} y^{2} - \frac{5}{6} y + 2 x + \frac{1}{6} = 0

in the standard form

4 (- \frac{3}{2}) (x - \frac{17}{96}) = (y - \frac{5}{4})^{2}

Therefore parabola properties are:

(h, k) = (\frac{17}{96}, \frac{5}{4}), p = - \frac{3}{2}

x^{2} - 16 y = 0

4 x^{2} + 48 y + 12 x - 159 = 0

y - 3 x^{2} + 10 x = 5

x, 2 x^{2} + 5 x + 2 = - 3 (2 w + 1)

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

What is 1/3 y^2-5/6 y+2x+1/6 =0 ?
The solution to 1/3 y^2-5/6 y+2x+1/6 =0 is Parabola with (h,k)=(17/96 , 5/4),p=-3/2