What is ((y-5)^2)/4 =(40(x+0))/4 ?

The solution to ((y-5)^2)/4 =(40(x+0))/4 is Parabola with (h,k)=(0,5),p=10

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((y-5)^2)/4 =(40(x+0))/4

\frac{( y - 5 ) ^{2}}{4} = \frac{40 ( x + 0 )}{4}

(h, k) = (0, 5), p = 10

Solution steps

\frac{( y - 5 ) ^{2}}{4} = \frac{40 ( x + 0 )}{4}

Rewrite

\frac{( y - 5 ) ^{2}}{4} = \frac{40 ( x + 0 )}{4}

in the standard form:

4 \cdot 10 (x - 0) = (y - 5)^{2}

Therefore parabola properties are:

(h, k) = (0, 5), p = 10

x^{2} - x - k (x) = 0

10 x^{2} - 256 = 300 f (x) in d

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

x^{2} - 10 - 15 y + 21 = 0

y = - \frac{1}{12} (x - 2)^{2} + 2

What is ((y-5)^2)/4 =(40(x+0))/4 ?
The solution to ((y-5)^2)/4 =(40(x+0))/4 is Parabola with (h,k)=(0,5),p=10