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whats a focal chord? I was givin this homework problem and instructor confused me on all the different positions
Any chord to y2 = 4ax which passes through the focus is called a focal chord of the parabola y2 = 4ax.
@lilsis76 tell me the focus of ur parabola?????:)
well, it didnt give me a focus it just gave me the P at (2,4)
it also says the parabola is y=x^2 so...um the p, 1/4
@lilsis76 focus of x2 = 4ay is(0,a) so compare it with x^2=y this implies 4a=1 or a=1/4 so it is(0,1/4).k
okay im back!
now you have 2 points p(2,4) and focus f(0,1/4) so can u draw a line???????????
yes i can
so thats ur focal chord any chord passing through focus is called focal chord:)
let me write that down
did i make the line right?
ya you make it right:) sry i didn't noticed ur reply:( @lilsis76
haha its okay. but thank u for letting me know. Okay whats next what do i have to do?
|dw:1352586259141:dw| now what is the coordinate of any point on parabola?????
wait, its a normal looking parabola with the point (0,0) right? thats how you got that parabolas shape?
im sorry im kinda new to this lol
since you said any point. ill go to the left. the Q would be P's opposite and be .. -2,4? right?
well it is not what i asked i asked general form it is(x,x^2) :) as PFQ is a straight line so slope of PF=FQ now find their slopes...........when u solve this completely i will tell you a short method:)
okay. im trying to do PF, is that ...getting the 2*1/4?
hmm... let me try again
WAIT let me try again
i got..... 4 - 1/4 / 2-0
|dw:1352587321957:dw| i dont know how to work out the fraction and whole number
slope m=|dw:1352587256238:dw| solve for x i just compare the two slopes:)
@lilsis76 is it value of x?????????
yes? how do i know?
i have to solve:(
oh. well then imma say yes it is then
how did you get the 13/8?
|dw:1352588075547:dw| so x=-1/8 or 2
haha okay. thanks
coordinates are (x,x^2) from fig x<0 x cannot be =2 Q=(-1/8,1/16)
okay i got the -1/8
and the 2, did you just add the 1/8 th twice to get 1/16?
now shortcut method:) Let y2 = 4ax be the equation of a parabola and (at^2, 2at) a point P on it. Suppose the coordinates of the other extremity Q of the focal chord through P are (at1^2, 2at1). then t1=-1/t:) so |dw:1352588333605:dw|
no @lilsis76 as the coordinates are(x,x^2)
@lilsis76 did u got it?????????
kinda, not really. okay well i ended up getting the - 1/8 for x, i plug it in right?
ok i got 8x+1 an x-2, x= -1/8, x=2
now go ahead for shortcut:)
the shortcut u gave me?
i got -1/8
anywho, i did the math cuz shortcut got me lost. i have x = -1/8 and x=2