Parabolas (pt2)

Q1

We have the form

( y - k)^2  =   4a ( x - h)

The vertex  is  given by  ( h, k)  =  (3, -1)

 Since  y is squared and we have no negatives on either side of the equation, it opens to the right

4a  = 12

a  = 3       which means that the focus is 3 units to the right of the vertex

The directrix  is 2a  = 2(3)   = 6 units from the focus

The focus  is at   ( 3 + 3, -1)   =   (6 - 1)

The dirctrix has the equation   x = (x coordinate of the vertex  - a)  =  (3 - 3)  =  0

So  x  = 0

Here's the graph :    https://www.desmos.com/calculator/jrejmuykh9

Q2

The directrix is to the left of the focus...so....this parabola opens to the right....therefore...we will hav no negatives on either side of the equation

The form will  be

(y - k)^2  = 4a (x  - h)

The vertex is given by   (  (sum of directrix + x coordinate of the focus ) / 2,   y coordinate of focus ) =

( [ 0 + 2]/2 , 6)  = (1, 6)  = (h, k)

"a"  is the distance from the vertex to the focus =  1

So we have

(y - 6)^2  = 4(1) ( x - 1)

( y - 6)^2 = 4 ( x - 1)

Here's the graph : https://www.desmos.com/calculator/1y3feymzzs