lilsis76
11.2 the Parabola. Its says Conic sections:
suppose PQ is a focal chord of the parabola y=x^2 and that the coordinates of P are (2,4)
a)find coordinate of Q
b)find coordinates of M, the midpoint of PQ.
c) A perpendicular is drawn from M to the y axis, meeting the y axis at S. Also, a line perpendicular to the focal chord is drawn through M, meeting the yaxis and T.
Find ST, and verify its equal to one-half the focal width of the parabola.
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lilsis76
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|dw:1352584620353:dw|
lilsis76
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|dw:1352584696851:dw|
lilsis76
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whats a focal chord? I was givin this homework problem and instructor confused me on all the different positions
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Any chord to y2 = 4ax which passes through the focus is called a focal chord of the parabola y2 = 4ax.
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@lilsis76 tell me the focus of ur parabola?????:)
lilsis76
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well, it didnt give me a focus it just gave me the P at (2,4)
lilsis76
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it also says the parabola is y=x^2 so...um
the p, 1/4
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@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
lilsis76
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okay im back!
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now you have 2 points p(2,4) and focus f(0,1/4)
so can u draw a line???????????
lilsis76
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yes i can
lilsis76
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|dw:1352585497197:dw|
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so thats ur focal chord
any chord passing through focus is called focal chord:)
lilsis76
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let me write that down
lilsis76
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did i make the line right?
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ya you make it right:)
sry i didn't noticed ur reply:(
@lilsis76
lilsis76
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haha its okay. but thank u for letting me know.
Okay whats next what do i have to do?
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|dw:1352586259141:dw|
now what is the coordinate of any point on parabola?????
lilsis76
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wait, its a normal looking parabola with the point (0,0) right? thats how you got that parabolas shape?
lilsis76
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im sorry im kinda new to this lol
lilsis76
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since you said any point. ill go to the left. the Q would be P's opposite and be .. -2,4? right?
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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:)
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@lilsis76
lilsis76
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okay. im trying to do PF, is that ...getting the 2*1/4?
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PF??????????
lilsis76
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hmm... let me try again
lilsis76
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1/4th?
lilsis76
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WAIT let me try again
lilsis76
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i got..... 4 - 1/4 / 2-0
lilsis76
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|dw:1352587321957:dw| i dont know how to work out the fraction and whole number
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slope m=|dw:1352587256238:dw|
solve for x
i just compare the two slopes:)
lilsis76
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15/4/2--> 15/8
lilsis76
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@lilsis76 is it value of x?????????
lilsis76
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yes? how do i know?
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i have to solve:(
lilsis76
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oh. well then imma say yes it is then
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|dw:1352587871193:dw|
lilsis76
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how did you get the 13/8?
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|dw:1352588075547:dw|
so x=-1/8 or 2
lilsis76
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haha okay. thanks
lilsis76
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ugh factoring
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coordinates are (x,x^2) from fig x<0
x cannot be =2
Q=(-1/8,1/16)
lilsis76
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okay i got the -1/8
lilsis76
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and the 2, did you just add the 1/8 th twice to get 1/16?
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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|
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no @lilsis76 as the coordinates are(x,x^2)
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@lilsis76 did u got it?????????
lilsis76
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kinda, not really. okay well i ended up getting the - 1/8 for x, i plug it in right?
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yup:)
lilsis76
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ok i got
8x+1 an x-2, x= -1/8, x=2
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now go ahead for shortcut:)
lilsis76
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the shortcut u gave me?
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yup:)
lilsis76
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i got -1/8
lilsis76
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lilsis76
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anywho, i did the math cuz shortcut got me lost.
i have x = -1/8 and x=2