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

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Answer 1 · 2016-05-21T03:18:09

Vertex V is at (0, 0), Focus S is at $(-1/4, 0)$ and the directrix is along $x = 1/4$ .

$y^2=- x > 0$ . So, $x<0$ .

The line of symmetry is y = 0.,

So, y = 0 is the axis.

As x < o, it is in the negative x-direction, from the vertex V..

V is (0, 0)

The latus rectum 4a = 1. So, a = 1/4).

The focus S is on the axism at a distance a from V.

So, S is at $(-1/4, 0)$

The directrix is perpendicular to the axis and is equidistant from the

V, on the opposite side of V. So, its equation is $x = 1/4$ .

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