What is y^2-8x+2y+17=0 ?

The solution to y^2-8x+2y+17=0 is Parabola with (h,k)=(2,-1),p=2

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y^2-8x+2y+17=0

y^{2} - 8 x + 2 y + 17 = 0

(h, k) = (2, - 1), p = 2

Solution steps

y^{2} - 8 x + 2 y + 17 = 0

Rewrite

y^{2} - 8 x + 2 y + 17 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (2, - 1), p = 2

x^{2} = - 36 y

\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1

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

\frac{x ^{2}}{16} - \frac{y ^{2}}{36} = 1

x^{2} + y^{2} - 2 x + 6 y + 9 = 0

What is y^2-8x+2y+17=0 ?
The solution to y^2-8x+2y+17=0 is Parabola with (h,k)=(2,-1),p=2