What is (x+1)^2+(y+3^2)=5 ?

The solution to (x+1)^2+(y+3^2)=5 is Parabola with (h,k)=(-1,-4),p=-1/4

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(x+1)^2+(y+3^2)=5

(x + 1)^{2} + (y + 3^{2}) = 5

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

Solution steps

(x + 1)^{2} + (y + 3^{2}) = 5

Rewrite

(x + 1)^{2} + (y + 3^{2}) = 5

in the standard form:

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

Therefore parabola properties are:

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

3 x^{2} - 13 y + 12 = 0

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

(5 x - 4) = \frac{y}{5 x - 4}

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

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

What is (x+1)^2+(y+3^2)=5 ?
The solution to (x+1)^2+(y+3^2)=5 is Parabola with (h,k)=(-1,-4),p=-1/4