What is x^2+5x-1/4 y+6=0 ?

The solution to x^2+5x-1/4 y+6=0 is Parabola with (h,k)=(-5/2 ,-1),p= 1/16

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x^2+5x-1/4 y+6=0

x^{2} + 5 x - \frac{1}{4} y + 6 = 0

(h, k) = (- \frac{5}{2}, - 1), p = \frac{1}{16}

Solution steps

x^{2} + 5 x - \frac{1}{4} y + 6 = 0

Rewrite

x^{2} + 5 x - \frac{1}{4} y + 6 = 0

in the standard form

4 \cdot \frac{1}{16} (y - (- 1)) = (x - (- \frac{5}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{5}{2}, - 1), p = \frac{1}{16}

1 - x^{2} - y = 0

x^{2} - 2 y + 1 = 0

v^{2} = v^{2} + 2 a s

(x + 1)^{2} = 32 (y - 0)

2 x^{2} + 7 y = 0

What is x^2+5x-1/4 y+6=0 ?
The solution to x^2+5x-1/4 y+6=0 is Parabola with (h,k)=(-5/2 ,-1),p= 1/16