What is x^2+(y+4)=2sqrt(2) ?

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

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x^2+(y+4)=2sqrt(2)

x^{2} + (y + 4) = 22

(h, k) = (0, - 4 + 22), p = - \frac{1}{4}

Solution steps

x^{2} + (y + 4) = 22

Rewrite

x^{2} + (y + 4) = 22

in the standard form:

4 (- \frac{1}{4}) (y - (- 4 + 22)) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, - 4 + 22), p = - \frac{1}{4}

x^{2} = \frac{64}{3} (3 - y)

1 = x - 0^{2} + (y - 1)^{2}

\frac{( x + 2 )}{16} + \frac{( y - 3 ) ^{2}}{49} = 1

y = - 0.005 (x)^{2} + 190

y^{2} - x - 16 = 0

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