Question

Screenshot attached.

Find focus and directrix of each parabola in figure.

Answer 1

Based solely on the graph? D:

Answer 2

Well yeah loool :P . I just picked up Cal-2 as a course and these 6 class stupid questions popped up. Should i just say that Focus is (p,0) and directrix is X=-p? :P Loool

Answer 3

Good thing these graphs are pretty easy.

The one on the left is horizontal, so its equation would be of the form
\[\Large x = a(y-k)^2+h\]

Answer 4

Where (h,k) is the vertex.
Thankfully, the vertex happens to be (0,0)
That certainly simplifies things.
\[\Large x = ay^2\]
Now you just have to figure out the value of a.

Answer 5

Wouldn't the equation be y^2=4px :/

Answer 6

Well, if you like.
\[\Large y^2 = 4px\]
Still, you could figure out p from here, I guess? :D

Answer 7

How am i supposed to do that?

Answer 8

With magic.
OR just recognising the fact that the graph crosses the point (1,1) :P
\[\Large \color{blue}1^2=4p(\color{red}1)^2\]

Answer 9

So.... that's done, then? :D

Answer 10

I'll just ask my little brother. Thanks :)

Answer 11

But it's already practically solved D:
\[\Large 1 = 4p\]

Answer 12

Well how can we assume it to be 1? Oh! okay :p

Answer 13

We don't... we just know that the graph passes the point (1,1)
So that means (1,1) must 'fit' into the equation
\[\Large y^2 = 4px\]

Answer 14

Speaking of which, I made an error in that earlier remark, it should have been
\[\Large \color{blue}1^2=4p(\color{red}1)\]

Answer 15

You sure man because everything is out of my mind :p

Answer 16

I am sure.

Take it from me, if the graph of an equation involving x and y crosses a point (a,b), then the equation MUST hold for x = a and y = b.

Answer 17

Same goes for the second parabola, except it's vertical this time, and so we have an equation
\[\Large x^2 = 4py\] which passes through the point (3,-3)
Meaning we get
\[\Large (\color{red}{-3})^2= 4p(\color{blue}3)\] and then solve for THIS p.

Answer 18

That's much like it then.Thanks alot :) .
*Is Cal-2 all about geometry?*

Answer 19

Whoops, backwards... my bad...
\[\Large (\color{red}{3})^2= 4p(\color{blue}{-3})\]

Answer 20

I don't know what Cal-2 is... all I see is THIS problem you've presented.

But if you suspect so, then better brush up on coordinate geom ^.^

Answer 21

Well yeah i was supposed to be the best on it but i haven't done it for like 4 years so yeah i need to brush up on it and i'll want you to help me through :D

Answer 22

I'm not the best at TEACHING geometry, though.
Got any more questions? :D

Answer 23

Lots of them but i'll get back to you later :)

Answer 24

I have a quiz tomorrow and the teacher supposes we know everything,well we do but i need to do all the working :p

Answer 25

Best of luck :P

Answer 26

@terenzreignz Well now i want you to explain me by drawing the whole thing as i ain't getting it :( Sorry :P