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

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

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

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

(h, k) = (- 2, 2), p = - 1

Solution steps

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

Rewrite

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

in the standard form

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

Therefore parabola properties are:

(h, k) = (- 2, 2), p = - 1

x^{2} + 8 y = 0

\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1

3 x^{2} + 3 y^{2} = 75

x^{2} + 4 y^{2} + 6 x + 16 y + 21 = 0

y^{2} = 12 x

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