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.

- lilsis76

- schrodinger

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- lilsis76

|dw:1352584620353:dw|

- lilsis76

|dw:1352584696851:dw|

- lilsis76

whats a focal chord? I was givin this homework problem and instructor confused me on all the different positions

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## More answers

- anonymous

Any chord to y2 = 4ax which passes through the focus is called a focal chord of the parabola y2 = 4ax.

- anonymous

@lilsis76 tell me the focus of ur parabola?????:)

- lilsis76

well, it didnt give me a focus it just gave me the P at (2,4)

- lilsis76

it also says the parabola is y=x^2 so...um
the p, 1/4

- anonymous

@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

okay im back!

- anonymous

now you have 2 points p(2,4) and focus f(0,1/4)
so can u draw a line???????????

- lilsis76

yes i can

- lilsis76

|dw:1352585497197:dw|

- anonymous

so thats ur focal chord
any chord passing through focus is called focal chord:)

- lilsis76

let me write that down

- lilsis76

did i make the line right?

- anonymous

ya you make it right:)
sry i didn't noticed ur reply:(
@lilsis76

- lilsis76

haha its okay. but thank u for letting me know.
Okay whats next what do i have to do?

- anonymous

|dw:1352586259141:dw|
now what is the coordinate of any point on parabola?????

- lilsis76

wait, its a normal looking parabola with the point (0,0) right? thats how you got that parabolas shape?

- lilsis76

im sorry im kinda new to this lol

- lilsis76

since you said any point. ill go to the left. the Q would be P's opposite and be .. -2,4? right?

- anonymous

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:)

- anonymous

@lilsis76

- lilsis76

okay. im trying to do PF, is that ...getting the 2*1/4?

- anonymous

PF??????????

- lilsis76

hmm... let me try again

- lilsis76

1/4th?

- lilsis76

WAIT let me try again

- lilsis76

i got..... 4 - 1/4 / 2-0

- lilsis76

|dw:1352587321957:dw| i dont know how to work out the fraction and whole number

- anonymous

slope m=|dw:1352587256238:dw|
solve for x
i just compare the two slopes:)

- lilsis76

15/4/2--> 15/8

- lilsis76

@Aperogalics

- anonymous

@lilsis76 is it value of x?????????

- lilsis76

yes? how do i know?

- anonymous

i have to solve:(

- lilsis76

oh. well then imma say yes it is then

- anonymous

|dw:1352587871193:dw|

- lilsis76

how did you get the 13/8?

- anonymous

|dw:1352588075547:dw|
so x=-1/8 or 2

- lilsis76

haha okay. thanks

- lilsis76

ugh factoring

- anonymous

coordinates are (x,x^2) from fig x<0
x cannot be =2
Q=(-1/8,1/16)

- lilsis76

okay i got the -1/8

- lilsis76

and the 2, did you just add the 1/8 th twice to get 1/16?

- anonymous

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|

- anonymous

no @lilsis76 as the coordinates are(x,x^2)

- anonymous

@lilsis76 did u got it?????????

- lilsis76

kinda, not really. okay well i ended up getting the - 1/8 for x, i plug it in right?

- anonymous

yup:)

- lilsis76

ok i got
8x+1 an x-2, x= -1/8, x=2

- anonymous

now go ahead for shortcut:)

- lilsis76

the shortcut u gave me?

- anonymous

yup:)

- lilsis76

i got -1/8

- lilsis76

@Aperogalics

- lilsis76

anywho, i did the math cuz shortcut got me lost.
i have x = -1/8 and x=2

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