How do you find the vertex, focus and directrix of $- 48 x = y^{2}$ $-48x = y^2$ ?

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Answer 1 · 2016-09-08T13:12:39

(1) the Vertex is, $(0, 0)$ $(0,0)$ .

(2) Focus is $(a, 0) = (- 12, 0)$ $(a,0)=(-12,0)$ , and,

(3) Eqn. of Directrix is $: x + a = 0, i . e ., x = 12$ $: x+a=0, i.e., x=12$ .

Comparing with the Standard Eqn. $y^{2} = 4 a x, a = - 12$ $y^2=4ax, a=-12$ .

Hence, (1) the Vertex is, $(0, 0)$ $(0,0)$ .

(2) Focus is $(a, 0) = (- 12, 0)$ $(a,0)=(-12,0)$ , and,

(3) Eqn. of Directrix is $: x + a = 0, i . e ., x = 12$ $: x+a=0, i.e., x=12$ . graph{y^2=-48x [-118.6, 118.55, -59.2, 59.3]}

How do you find the vertex, focus and directrix of -48x = y^2 −48x=y2-48x = y^2 ?