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

The solution to 4x^2+4^2+4x+4y+4=0 is Parabola with (h,k)=(-1/2 ,-19/4),p=-1/4

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

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

(h, k) = (- \frac{1}{2}, - \frac{19}{4}), p = - \frac{1}{4}

Solution steps

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

Rewrite

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

in the standard form

4 (- \frac{1}{4}) (y - (- \frac{19}{4})) = (x - (- \frac{1}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{2}, - \frac{19}{4}), p = - \frac{1}{4}

(x + 7)^{2} = - 2 (y - 6)

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

x = 5 y^{2} + 10 y - 15

y^{2} + 6 y + 4 x - 3 = 0

h (x) = - (x - 11) (x + 3)

What is 4x^2+4^2+4x+4y+4=0 ?
The solution to 4x^2+4^2+4x+4y+4=0 is Parabola with (h,k)=(-1/2 ,-19/4),p=-1/4