What is x^2+y+2x-4y-4=0 ?

The solution to x^2+y+2x-4y-4=0 is Parabola with (h,k)=(-1,-5/3),p= 3/4

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x^2+y+2x-4y-4=0

x^{2} + y + 2 x - 4 y - 4 = 0

(h, k) = (- 1, - \frac{5}{3}), p = \frac{3}{4}

Solution steps

x^{2} + y + 2 x - 4 y - 4 = 0

Rewrite

x^{2} + y + 2 x - 4 y - 4 = 0

in the standard form

4 \cdot \frac{3}{4} (y - (- \frac{5}{3})) = (x - (- 1))^{2}

Therefore parabola properties are:

(h, k) = (- 1, - \frac{5}{3}), p = \frac{3}{4}

x^{2} + (y + 1) 2 = 36

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

x^{2} - 18 x + 24 y + 273 = 0

y + 2 = x^{2} + 8 x

\frac{2}{3} x \leq y^{2}

What is x^2+y+2x-4y-4=0 ?
The solution to x^2+y+2x-4y-4=0 is Parabola with (h,k)=(-1,-5/3),p= 3/4