hba
  • hba
Screenshot attached. Find focus and directrix of each parabola in figure.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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hba
  • hba
1 Attachment
terenzreignz
  • terenzreignz
Based solely on the graph? D:
hba
  • hba
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

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terenzreignz
  • terenzreignz
Good thing these graphs are pretty easy. The one on the left is horizontal, so its equation would be of the form \[\Large x = a(y-k)^2+h\]
terenzreignz
  • terenzreignz
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.
hba
  • hba
Wouldn't the equation be y^2=4px :/
terenzreignz
  • terenzreignz
Well, if you like. \[\Large y^2 = 4px\] Still, you could figure out p from here, I guess? :D
hba
  • hba
How am i supposed to do that?
terenzreignz
  • terenzreignz
With magic. OR just recognising the fact that the graph crosses the point (1,1) :P \[\Large \color{blue}1^2=4p(\color{red}1)^2\]
terenzreignz
  • terenzreignz
So.... that's done, then? :D
hba
  • hba
I'll just ask my little brother. Thanks :)
terenzreignz
  • terenzreignz
But it's already practically solved D: \[\Large 1 = 4p\]
hba
  • hba
Well how can we assume it to be 1? Oh! okay :p
terenzreignz
  • terenzreignz
We don't... we just know that the graph passes the point (1,1) So that means (1,1) must 'fit' into the equation \[\Large y^2 = 4px\]
terenzreignz
  • terenzreignz
Speaking of which, I made an error in that earlier remark, it should have been \[\Large \color{blue}1^2=4p(\color{red}1)\]
hba
  • hba
You sure man because everything is out of my mind :p
terenzreignz
  • terenzreignz
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.
terenzreignz
  • terenzreignz
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.
hba
  • hba
That's much like it then.Thanks alot :) . *Is Cal-2 all about geometry?*
terenzreignz
  • terenzreignz
Whoops, backwards... my bad... \[\Large (\color{red}{3})^2= 4p(\color{blue}{-3})\]
terenzreignz
  • terenzreignz
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 ^.^
hba
  • hba
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
terenzreignz
  • terenzreignz
I'm not the best at TEACHING geometry, though. Got any more questions? :D
hba
  • hba
Lots of them but i'll get back to you later :)
hba
  • hba
I have a quiz tomorrow and the teacher supposes we know everything,well we do but i need to do all the working :p
terenzreignz
  • terenzreignz
Best of luck :P
hba
  • hba
@terenzreignz Well now i want you to explain me by drawing the whole thing as i ain't getting it :( Sorry :P

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