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

The solution to (y+4)^2+(x-2)=16 is Parabola with (h,k)=(18,-4),p=-1/4

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

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

(h, k) = (18, - 4), p = - \frac{1}{4}

Solution steps

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

Rewrite

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

in the standard form:

4 (- \frac{1}{4}) (x - 18) = (y - (- 4))^{2}

Therefore parabola properties are:

(h, k) = (18, - 4), p = - \frac{1}{4}

What is (y+4)^2+(x-2)=16 ?
The solution to (y+4)^2+(x-2)=16 is Parabola with (h,k)=(18,-4),p=-1/4