SOLUTION: a parabola has a vertex v=(6, 4) and a focus f=(6, -1). Enter the equation in the form: (x-h)^2=4p(y-k) or (y-k)^2=4p(x-h)

Question 925550: a parabola has a vertex v=(6, 4) and a focus f=(6, -1). Enter the equation in the form: (x-h)^2=4p(y-k) or (y-k)^2=4p(x-h)
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
You are given enough information to find the directrix. Use the distance formula and definition of a parabola to derive the equation for the specific parabola of your example; and adjust the form of the equation to whichever format you want.

The directrix is on the other side of the vertex than the focus. The directrix is y=9, so the general point for the directrix would be (x,9).

Work with this:
Focus to parabola equals directrix to parabola.

RELATED QUESTIONS
A parabola has a Vertex of V=(6,4) and a focus of F=(6,-1). What is the equation?... (answered by ewatrrr)

Find an equation of the parabola that has vertex (1,2) and focus (-1,2). I think that... (answered by scott8148)

Find an equation in standard form of the parabola described. Vertex at (−2, 4); passes (answered by MathLover1)

a Parabola has a vertex (-1,5) and contains the point (2,-4). Write an equation of the... (answered by nerdybill)

Could someone please help me with writing a conic equation in standard form? The... (answered by josgarithmetic,Natolino1983)

Use the definition of a parabola to show that the parabola with the vertex(h,k) and focus (answered by scott8148)

Put the equation y=4x^2+7x+6 in vertex form and state the values of a, h, and k. Enter... (answered by stanbon)

Find an equation in standard form of the parabola described. Vertex at (−3, −3);... (answered by MathLover1)

Please explain why the graph of (x-h)^2 = 4p(y-k) has a y-intercept of (0, h^2/4p +K). (answered by Nate)