Question

Find the directrix, focus, and vertex of the parabola whose equation is: y^2-8y+116=-20x

Answer 1

this involves completing the square
you need to write this in the form
\[(y-k)^2=4p(x-h)\]

Answer 2

so first let's look at the y^2-8y+?
we need to figure out what ? should be so we can complete the square

Answer 3

16

Answer 4

perfect
y^2-8y+16=(y-4)^2

Answer 5

\[y^2-8y+116=-20x \\ y^2-8y+16+100=-20x \\ (y-4)^2+100=-20x\]
we also need to isolate the squared part

Answer 6

just in it is here:
\[(y-k)^2=4p(x-h)\]

Answer 7

if you need a hint to do this.. let me know

Answer 8

\[(y-4)^2=-20(x-5)\]

Answer 9

x+5

Answer 10

cool

Answer 11

\[(y-4)^2=-20(x+5)\]

Answer 12

Vertex: (-5,4)

Answer 13

yep

Answer 14

now in place of 4p we see -20
so p is ?

Answer 15

-5

Answer 16

right

|p| is the distance between the vertex and the directrix
or the distance between the vertex and the focus

(since the square is on the y stuff)
also the - part means the parabola is opened to the left
if it was + it would be opened to the right


------
(if the square was on the x stuff)
- means down
+ means up

Answer 17

so the focus is (-5,9)?

Answer 18

you chanced the wrong coordinate