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

Question

Lincoln Anderson · Accepted Answer

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

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

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

Comparing with the Standard Eqn. # y^2=4ax, a=-12#. Hence, (1) the Vertex is, #(0,0)#. (2) Focus is #(a,0)=(-12,0)#, and, (3) Eqn. of Directrix is # : x+a=0, i.e., x=12#. graph{y^2=-48x [-118.6, 118.55, -59.2, 59.3]}

Chloe Alexander · Answer

undefined To find the vertex, focus, and directrix of the given equation $ -48x = y^2 $, you can first rewrite it in the standard form of a parabola equation $ y^2 = 4px $. Comparing the given equation with the standard form, we have $ 4p = -48 $, which implies $ p = -12 $.

1. **Vertex**: The vertex of the parabola is at the origin since the equation is in the form $ y^2 = 4px $. Therefore, the vertex is $ (0, 0) $.

2. **Focus**: The focus of the parabola is given by the point $ (p, 0) $. Substituting $ p = -12 $, we find the focus at $ (-12, 0) $.

3. **Directrix**: The directrix of the parabola is a vertical line parallel to the y-axis, located at a distance of $ -p $ units from the vertex. Thus, the directrix is the line $ x = -(-12) = 12 $.