Question

Identify the vertex, focus, and directrix. y=1/8y^2

Answer 1

If all the equations for the directrix are "x = " lines then this is a y^2 parabola.  The actual equation is $x= \frac{1}{8}y^2$.  The standard form for a positive sideways-opening parabola is  $(y-k)^2=4p(x-h)$.  We know from the equation that the vertex of the parabola is at the origin, or else the translation would be reflected within the parenthesis in the equation.  Our equation has no parenthesis to indicate movement from the origin.  The vertex is (0, 0).  Got that out of the way.  That simplifies our standard form down to  $y^2=4p(x)$.  Let's take a look at our equation now.  It is  $x= \frac{1}{8}y^2$.  We could rewrite it and make it a closer match to the standard form if we multiply both sides by 8 to get rid of the fraction.  That gives us an equation that looks like this:  $y^2=8x$.  That means that 4p = 8, and p = 2.  p is the distance that the focus and the directrix are from the vertex.  Since this is a positive parabola, it opens up to the right.  Which means, then, that the focus is to the right of the vertex, 2 units to be exact, and the directrix is 2 units to the left of the vertex.  The formula for the focus is (h + p, k).  Our h is 0, our k is 0 and our p is 2, so the coordinates of the focus are (2, 0).  Going 2 units to the left of the origin then puts our directrix at the line x = -2.  Your choice then as your answer is b.