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

The solution to 3y^2-x+3y+6=0 is Parabola with (h,k)=(21/4 ,-1/2),p= 1/12

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3y^2-x+3y+6=0

3 y^{2} - x + 3 y + 6 = 0

(h, k) = (\frac{21}{4}, - \frac{1}{2}), p = \frac{1}{12}

Solution steps

3 y^{2} - x + 3 y + 6 = 0

Rewrite

3 y^{2} - x + 3 y + 6 = 0

in the standard form

4 \cdot \frac{1}{12} (x - \frac{21}{4}) = (y - (- \frac{1}{2}))^{2}

Therefore parabola properties are:

(h, k) = (\frac{21}{4}, - \frac{1}{2}), p = \frac{1}{12}

x^{2} - 4 x + 8 y - 12 = 0

\frac{y}{x} = \frac{y}{x + 15} + \frac{1}{3}

y^{2} - 6 y - 12 x + 33 = 0

- 3 x^{2} + 12 x - y - 9 = 0

3 y = 21 x^{2} + 9 x + 6

What is 3y^2-x+3y+6=0 ?
The solution to 3y^2-x+3y+6=0 is Parabola with (h,k)=(21/4 ,-1/2),p= 1/12