What is 16x+16y^2+8x-32y=-1 ?

The solution to 16x+16y^2+8x-32y=-1 is Parabola with (h,k)=(5/8 ,1),p=-3/8

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16x+16y^2+8x-32y=-1

16 x + 16 y^{2} + 8 x - 32 y = - 1

(h, k) = (\frac{5}{8}, 1), p = - \frac{3}{8}

Solution steps

16 x + 16 y^{2} + 8 x - 32 y = - 1

Rewrite

16 x + 16 y^{2} + 8 x - 32 y = - 1

in the standard form

4 (- \frac{3}{8}) (x - \frac{5}{8}) = (y - 1)^{2}

Therefore parabola properties are:

(h, k) = (\frac{5}{8}, 1), p = - \frac{3}{8}

9 y = (\frac{12 x}{3 y}) + 3 y

y^{2} - 4 x + 6 y + 13 = 0

y^{2} + 12 x - 8 + 16 = 0

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

(y - 7)^{2} + 24 (x + 4) = 0

What is 16x+16y^2+8x-32y=-1 ?
The solution to 16x+16y^2+8x-32y=-1 is Parabola with (h,k)=(5/8 ,1),p=-3/8