What is 2x^2+2y-8x+16y-32=0 ?

The solution to 2x^2+2y-8x+16y-32=0 is Parabola with (h,k)=(2, 20/9),p=-9/4

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2x^2+2y-8x+16y-32=0

2 x^{2} + 2 y - 8 x + 16 y - 32 = 0

(h, k) = (2, \frac{20}{9}), p = - \frac{9}{4}

Solution steps

2 x^{2} + 2 y - 8 x + 16 y - 32 = 0

Rewrite

2 x^{2} + 2 y - 8 x + 16 y - 32 = 0

in the standard form

4 (- \frac{9}{4}) (y - \frac{20}{9}) = (x - 2)^{2}

Therefore parabola properties are:

(h, k) = (2, \frac{20}{9}), p = - \frac{9}{4}

y^{2} = 2 c x + 4

x 2 x = 5 y

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

2 x^{2} + 8 x - 3 + y = 0

\frac{d}{d X} (\frac{X ^{2}}{2} + 2 X - 3)

What is 2x^2+2y-8x+16y-32=0 ?
The solution to 2x^2+2y-8x+16y-32=0 is Parabola with (h,k)=(2, 20/9),p=-9/4