How do you find the vertex, focus and directrix of $x+y^2=0$ ?

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Answer 1 · 2018-08-03T12:42:16

Vertex is at $(0,0)$ , focus is at $(-0.25,0)$ and
directrix is $x= 0.25$

$x+y^2=0 or y^2 = -x or (y-0)^2 = -4 *1/4 (x-0)$

The equation of horizontal parabola opening left is

$(y-k)^2 = -4 p(x-h) ; h=0 ,k=0 ; p= 1/4=0.25$

Therefore, vertex is at $(0,0)$ The distance between focus and

vertex is $p=0.25 :.$ Focus is at $(-0.25,0)$ . Vertex is at midway

between focus and directrix. Therefore , directrix is $x= 0.25$

graph{x+y^2=0 [-12.66, 12.65, -6.33, 6.33]} [Ans]

How do you find the vertex, focus and directrix of x+y^2=0x+y^2=0?