+0

# Parabolas (pt2)

0
68
2

Sorry, these are really hard for me.

Q1 Q2 Nov 19, 2019

#1
+1

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   Nov 19, 2019
#2
+1

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   Nov 19, 2019