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I 4got how to get the equation of a parabola when given 3 points. I have (0,0) & (5.5,9) & (-5.5,9) how do I get this into ax^2+ bx + c which is the parabolic equation I believe
maybe we just substitute and solve - 3 equations 3 unknowns ?
ax^2+bx+c goes thru (0,0) a(0)^2+b(0)+c = 0 c = 0 ------------------(1)
so c-0? that means then the foci would be (0, 1/4) right?
hmm im not sure if we can figure out focus yet, how did u get that ? :)
Oh whoops used a different thing ok so on my lesson it says foci can be found by doing (h, k+c) Oh and if this helps vertex is 0,0
ohk :) vertex is (0, 0 ) yes.. cuz 5.5 & -5.5 give same value 9. and they are at same distance from y axis
that gives h = 0, k = 0
And c is 0?
we got that, focus = (h, k+1/4a)
K so (0,1/4a) how do we solve for a?
c = 0 is first equation, since it goes thru (5.5,9), a(5.5)^2+b(5.5)+0 = 9 -------------(2) since it gors thru (-5.5, 9), a(-5.5)^2+b(-5.5)+0 = 9 -------------(3)
solving (1), (2), (3) gives, a = 0.2975 b = 0 c = 0 so, the required parabola is y = 0.2975x^2
Well a=1/4c and hey why do u put a bunch of dashes then a number in parenthesis next to it?
ha that works :)
just to label it as equation (2), equation (3) etc...
Oh ok :)
so focus wud be (h, k+1/4a) = (0, 0+1/4(0.2975))
Wow nice yeah parabola intersects the points and k foci is (0,.84) approximate
yup (0, .84) is focus, so that gives directrix as y = -.84 it all boild down to solving system of equations i think
ok i see :p yeah I'm doing a conic section solar cookerr project thing & 4got how 2 find equation and foci of parabola, thx a ton :)
wow parabolic mirrors... hows it going ?
Need to poke something through the foci now and once I'm done with it I have to cook something with it :p
ive tried it using my dish antenna once.. it was a fail lol... couldnt complete it :(
oh is that for fun.. or wat ? :)
Lol that sucks :D after this then I can take my finals for algebra II
ohk... nice :) good luck wid the project :))