How do you find the vertex, focus, and directrix of the parabola $(x + 3) + {(y - 2)}^{2} = 0$ $(x+3)+(y-2)^2=0$ ?

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Answer 1 · 2018-07-01T12:54:28

Vertex is at $(- 3, 2)$ $(-3,2)$ , directrix is $x = - 2.75$ $x = - 2.75$ and
focus is at $(- 3.25, 2)$ $(-3.25,2)$

$(x + 3) + {(y - 2)}^{2} = 0 or {(y - 2)}^{2} = - (x + 3)$ $(x+3)+(y-2)^2=0 or (y-2)^2=-(x+3)$ or

${(y - 2)}^{2} = - 4 \cdot 0.25 (x + 3)$ $(y-2)^2=-4*0.25(x+3)$ The equation of horizontal

parabola opening left is

${(y - k)}^{2} = - 4 p (x - h); h = - 3, k = 2, p = 0.25; h, k$ $(y-k)^2 = -4p(x-h) ; h=-3 ,k=2 , p = 0.25 ; h ,k$

being vertex .Therefore vertex is at $(- 3, 2)$ $(-3,2)$

Distance between focus and vertex is $p = 0.25$ $p=0.25$

Here the directrix is at right of the vertex. Therefore the

equation of directrix is $x = (- 3 + 0.25) or x = - 2.75$ $x=(-3+0.25) or x = - 2.75$

Vertex is at midway between focus and directrix. Therefore,

focus is at $(- 3 - 0.25), 2 or (- 3.25, 2)$ $(-3-0.25),2 or (-3.25,2)$

graph{(x+3)+(y-2)^2=0 [-10, 10, -5, 5]} [Ans]

How do you find the vertex, focus, and directrix of the parabola (x+3)+(y-2)^2=0(x+3)+(y−2)2=0(x+3)+(y-2)^2=0?