What is 8y^2+64y+32x-48=0 ?

The solution to 8y^2+64y+32x-48=0 is Parabola with (h,k)=(11/2 ,-4),p=-1

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8y^2+64y+32x-48=0

8 y^{2} + 64 y + 32 x - 48 = 0

(h, k) = (\frac{11}{2}, - 4), p = - 1

Solution steps

8 y^{2} + 64 y + 32 x - 48 = 0

Rewrite

8 y^{2} + 64 y + 32 x - 48 = 0

in the standard form

4 (- 1) (x - \frac{11}{2}) = (y - (- 4))^{2}

Therefore parabola properties are:

(h, k) = (\frac{11}{2}, - 4), p = - 1

y^{2} + x - 9 = 0

y - 3 x^{2} + 2 x - 36 \leq 0

x^{(2)} - 10 x + 8 y + 33 = 0

x^{2} + y = 121

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

What is 8y^2+64y+32x-48=0 ?
The solution to 8y^2+64y+32x-48=0 is Parabola with (h,k)=(11/2 ,-4),p=-1