Question

How can I find the vertex, focus, directrix and focal width of a parabola with the equation -1/20x^2=y?

Answer 1

do you know what each of those things are...?

Answer 2

Yes! How do I determine them from the equation though?

Answer 3

so your vertex is the turning point of your graph, yeah?

Answer 4

Yes.. It would be at the origin in this case

Answer 5

so how did you want to work it out... you could either draw the graph and get your answers from there, or you can do it mathematically

Answer 6

Id like to do it mathematically, so that I can do it without a calculator in case I ever have to. If you could help me figure "P" out then I think I can do it by myself.

Answer 7

and your equation is...?
-(1/20) x^2=y
or
-1/(20x^2) =y

???

Answer 8

The first

Answer 9

k, well that can be re-written as -x^2 / 20 = y
so the turning poing of a parabola occurs when the gradient = 0
so what you can do is take the derivative (which give the gradient at any x point) and sub in y=0

Answer 10

so derivative is y = -x/10
so sub in 0
so 0 = -x/10
so x = 0
turning point is a 0,0

Answer 11

So when substitute for that, the answer will be P? In my class our formula for a vertical parabola is x^2=4py

Answer 12

...never heard of P, sorry

Answer 13

what does it let u work out?

Answer 14

Basically everything. For example, if p>0 and the parabola opens downward, then the focus is (h,k+p) and the directrix is y=k-p

Answer 15

Oh! Okay, thank you

Answer 16

so your focus is (0, -4) ...i think

Answer 17

you start with
\[ -\frac{1}{20} x^2=y \]
multiply both sides by -20 :
\[ -20 \cdot -\frac{1}{20} x^2= -20y \\ x^2 = -20y\]
factor out a 4 from the 20 (i.e. write 20 as 4 * 5):
\[ x^2 = 4(-5)y \]
as you can see p=-5

Answer 18

Thank you guys!

Answer 19

for the focus,
switch the terms so you have x = (something something) y
in this case x^2 = -20y
so x^2 = 4 *(-5)y
P = -5 (sorry, typo before)

Answer 20

No problem!

Answer 21

oooh, shiny!