What is y^2+4y+8x-12=0 ?

The solution to y^2+4y+8x-12=0 is Parabola with (h,k)=(2,-2),p=-2

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

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

(h, k) = (2, - 2), p = - 2

Solution steps

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

Rewrite

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

in the standard form

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

Therefore parabola properties are:

(h, k) = (2, - 2), p = - 2

13 x^{2} + 49 y^{2} = 637

4 x^{2} + 9 y^{2} = 36

x^{2} + \frac{y ^{2}}{16} = 1

y^{2} + 2 x^{2} = 1

\frac{( y + 3 ) ^{2}}{9} - \frac{( x - 2 ) ^{2}}{16} = 1

What is y^2+4y+8x-12=0 ?
The solution to y^2+4y+8x-12=0 is Parabola with (h,k)=(2,-2),p=-2