What is 16(x+2)-(y+5)^2=0 ?

The solution to 16(x+2)-(y+5)^2=0 is Parabola with (h,k)=(-2,-5),p=4

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16(x+2)-(y+5)^2=0

16 (x + 2) - (y + 5)^{2} = 0

(h, k) = (- 2, - 5), p = 4

Solution steps

16 (x + 2) - (y + 5)^{2} = 0

Rewrite

16 (x + 2) - (y + 5)^{2} = 0

in the standard form:

4 \cdot 4 (x - (- 2)) = (y - (- 5))^{2}

Therefore parabola properties are:

(h, k) = (- 2, - 5), p = 4

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

2 y^{2} - 10 x + 5 = 0

y^{2} - 4 y + 3 = x

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

x = 4 (y - 1)^{2} + 5

What is 16(x+2)-(y+5)^2=0 ?
The solution to 16(x+2)-(y+5)^2=0 is Parabola with (h,k)=(-2,-5),p=4