What are the vertex focus and directrix of a parabola with equation x=y^2+14y-2

This is a sideways opening parabola, opening to the right to be more specific, since the leading coefficient is a positive 1.  The rule for a focus and a directrix is that they are the same number of units from the vertex (in other words, the vertex is dead center between them), and that the vertex is on the same axis that the focus is.  We need to find the vertex then to determine what the focus and the directrix are.  We will complete the square on that to find the vertex.  Begin by setting it equal to 0, then move the 2 over by addition to get .  Now we will complete the square on the y terms.  Take half the linear term, square it, and add it to both sides.  Our linear term is 14.  Half of 14 is 7, and 7 squared is 49. So we add 49 to both sides. , which of course simplifies to .  The purpose of this is to find the k coodinate of the vertex which will be revealed when we write the perfect square binomial we created during this process: .  Moving the 51 back over by subtraction gives us .  The vertex then is (-51,-7).  The formula to find the focus using this vertex is .  As I stated quite a while ago, the leading coefficient on our parabola was a +1 so our "a" value is 1, and the focus is then found in which simplifies to .  If the vertex is (-51, -7) and the focus is (-50.75, -7), then the distance between them is 1/4, or .25.  That means that the directrix is also .25 units from the vertex, but in the other direction.  Our directrix is a vertical line, and it will have the equaion x = -51.25.  Summing up, your focus is (-50.75, -7) and your directrix is x = -51.25

The correct answer is -20y

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