What is (x-3)^2= 1/2 (y+1) ?

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

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(x-3)^2= 1/2 (y+1)

(x - 3)^{2} = \frac{1}{2} (y + 1)

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

Solution steps

(x - 3)^{2} = \frac{1}{2} (y + 1)

Rewrite

(x - 3)^{2} = \frac{1}{2} (y + 1)

in the standard form:

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

Therefore parabola properties are:

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

- x^{2} + 10 x + 3 y - 37 = 0

y^{2} = C x + \frac{1}{8} C^{3}

x = 7 y^{2} + 4

y^{2} + 6 y + 8 x + 1 = 0

x^{2} + 14 x - 8 y + 65 = 0

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