How do you find the vertex, focus and directrix of $x = - \frac{1}{2} {(y - 2)}^{2} - 4$ $x =-1/2(y-2)^2-4$ ?

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Answer 1 · 2018-04-09T12:57:02

Vertex is at $(- 4, 2)$ $(-4,2)$ , focus is at $(- 4.5, 2)$ $(-4.5,2)$ and
directrix is $x = - 3.5$ $x=-3.5$

$x = - \frac{1}{2} {(y - 2)}^{2} - 4 or - \frac{1}{2} {(y - 2)}^{2} = x + 4$ $x=-1/2(y-2)^2-4 or -1/2(y-2)^2=x+4$ or

${(y - 2)}^{2} = - 2 (x + 4) or {(y - 2)}^{2} = - 4 \cdot \frac{1}{2} (x + 4)$ $(y-2)^2=-2(x+4) or (y-2)^2=-4*1/2(x+4)$

Comparing with the equation of horizontal

parabola opening left is ${(y - k)}^{2} = - 4 p (x - h)$ $(y-k)^2 = -4p(x-h)$

We get , $h=-4 ,k=2, p=1/2 :.$ vertex is at

$(h,k) or (-4,2)$ , focus is at $(-4-1/2),2or (-4.5,2)$

and directrix is $x= (-4+1 /2) or x = -3.5$

graph{x= -1/2(y-2)^2-4 [-10, 10, -5, 5]}

How do you find the vertex, focus and directrix of x =-1/2(y-2)^2-4x=−12(y−2)2−4x =-1/2(y-2)^2-4?