What is (x-8)^2+(y-4)=16 ?

The solution to (x-8)^2+(y-4)=16 is Parabola with (h,k)=(8,20),p=-1/4

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(x-8)^2+(y-4)=16

(x - 8)^{2} + (y - 4) = 16

(h, k) = (8, 20), p = - \frac{1}{4}

Solution steps

(x - 8)^{2} + (y - 4) = 16

Rewrite

(x - 8)^{2} + (y - 4) = 16

in the standard form:

4 (- \frac{1}{4}) (y - 20) = (x - 8)^{2}

Therefore parabola properties are:

(h, k) = (8, 20), p = - \frac{1}{4}

x = - \frac{y ^{2}}{32}

y^{2} + 5 y + \frac{25}{4} = 8 x - 13 + \frac{25}{4}

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

(y + 1)^{2} = 2 (x + 1)

y \geq - 2 x^{2} - 2

What is (x-8)^2+(y-4)=16 ?
The solution to (x-8)^2+(y-4)=16 is Parabola with (h,k)=(8,20),p=-1/4