Question

how do i find the directrix and focus of this:
(y-2)^2=8(x+1) please help! my test is tomorrow and I have no idea how to do it.

Answer 1

do you know where the vertex of this parabola is ???

Answer 2

do you if this is a parabola opening vertically or sideways??

Answer 3

parabolas that open either up/down (vertically) have the equation: \(\large (y-k)^2=4p(x-h) \)

parabolas that open either left/right (sideways) have the equation: \(\large (x-h)^2=4p(y-k) \)

both parabolas have the vertex at (h, k).

so once you know where the vertex is and which way the parabola opens, you can find the directrix to be p units away from the vertex.

Answer 4

ok, so in this problem, the vertex is (-1,2)

Answer 5

ok... and which way does this parabola open ????

Answer 6

ummm....up? I know how to determine where it opens if I have the value of p. But I dont know how to find it in this problem.

Answer 7

no....
if the squared term is the y, it is a parabola opening left/right...
if the squared term is the x, it is a parabola opening up/down...

which variable is squared in your equation??? the x or the y???

Answer 8

y, so...it would be left of right...so, im guessing right?

Answer 9

correct...
now to determine if it opens to the left or to the right we have to determine p...

Answer 10

ok. how?

Answer 11

\(\large (y-2)^2=8(x+1) \)
\(\large (y-2)^2=4\cdot \color {red}2(x+1) \)

do you see that p=2 ???

Answer 12

ok yeah

Answer 13

and since p is positive, this parabola opens to the right....

Answer 14

ok

Answer 15

so, to find it I divide the coefficient by 4?