What is 2y^2+2y^2-16x+4y+20=0 ?

The solution to 2y^2+2y^2-16x+4y+20=0 is Parabola with (h,k)=(19/16 ,-1/2),p=1

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2y^2+2y^2-16x+4y+20=0

2 y^{2} + 2 y^{2} - 16 x + 4 y + 20 = 0

(h, k) = (\frac{19}{16}, - \frac{1}{2}), p = 1

Solution steps

2 y^{2} + 2 y^{2} - 16 x + 4 y + 20 = 0

Rewrite

2 y^{2} + 2 y^{2} - 16 x + 4 y + 20 = 0

in the standard form

4 \cdot 1 \cdot (x - \frac{19}{16}) = (y - (- \frac{1}{2}))^{2}

Therefore parabola properties are:

(h, k) = (\frac{19}{16}, - \frac{1}{2}), p = 1

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

x^{2} - 2 x - 24 y + 1 = 0

x^{2} - 10 x - 96 y = - 25 - 96

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

y^{2} + 3 x - 16 y + 65 = 0

What is 2y^2+2y^2-16x+4y+20=0 ?
The solution to 2y^2+2y^2-16x+4y+20=0 is Parabola with (h,k)=(19/16 ,-1/2),p=1