How do you find the vertex, focus, and directrix of the parabola $y^{2} - 4 y - 4 x = 0$ $y^2-4y-4x=0$ ?

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Answer 1 · 2017-01-04T12:56:41

The vertex is V $= (- 1, 2)$ $=(-1,2)$
The focus is F $= (0, 2)$ $=(0,2)$
The directrix is $x = - 2$ $x=-2$

Let's rearrange the equation and complete the squares

$y^{2} - 4 y = 4 x$ $y^2-4y=4x$

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

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

Comparing this equation to

${(y - b)}^{2} = 2 p (x - a)$ $(y-b)^2=2p(x-a)$

$p = 2$ $p=2$

The vertex is V $= (a, b) = (- 1, 2)$ $=(a,b)=(-1,2)$

The focus is F $= (a + \frac{p}{2}, b) = (0, 2)$ $=(a+p/2,b)=(0,2)$

The directrix is $x = a - \frac{p}{2}$ $x=a-p/2$

$x = - 1 - 1 = - 2$ $x=-1-1=-2$

graph{(y^2-4y-4x)(y-100x-200)=0 [-10, 10, -5, 5]}

How do you find the vertex, focus, and directrix of the parabola y^2-4y-4x=0y2−4y−4x=0y^2-4y-4x=0?