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lilsis76

  • 2 years ago

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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  1. lilsis76
    • 2 years ago
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    |dw:1352584620353:dw|

  2. lilsis76
    • 2 years ago
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    |dw:1352584696851:dw|

  3. lilsis76
    • 2 years ago
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    whats a focal chord? I was givin this homework problem and instructor confused me on all the different positions

  4. Aperogalics
    • 2 years ago
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    Any chord to y2 = 4ax which passes through the focus is called a focal chord of the parabola y2 = 4ax.

  5. Aperogalics
    • 2 years ago
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    @lilsis76 tell me the focus of ur parabola?????:)

  6. lilsis76
    • 2 years ago
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    well, it didnt give me a focus it just gave me the P at (2,4)

  7. lilsis76
    • 2 years ago
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    it also says the parabola is y=x^2 so...um the p, 1/4

  8. Aperogalics
    • 2 years ago
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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

  9. lilsis76
    • 2 years ago
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    okay im back!

  10. Aperogalics
    • 2 years ago
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    now you have 2 points p(2,4) and focus f(0,1/4) so can u draw a line???????????

  11. lilsis76
    • 2 years ago
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    yes i can

  12. lilsis76
    • 2 years ago
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    |dw:1352585497197:dw|

  13. Aperogalics
    • 2 years ago
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    so thats ur focal chord any chord passing through focus is called focal chord:)

  14. lilsis76
    • 2 years ago
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    let me write that down

  15. lilsis76
    • 2 years ago
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    did i make the line right?

  16. Aperogalics
    • 2 years ago
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    ya you make it right:) sry i didn't noticed ur reply:( @lilsis76

  17. lilsis76
    • 2 years ago
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    haha its okay. but thank u for letting me know. Okay whats next what do i have to do?

  18. Aperogalics
    • 2 years ago
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    |dw:1352586259141:dw| now what is the coordinate of any point on parabola?????

  19. lilsis76
    • 2 years ago
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    wait, its a normal looking parabola with the point (0,0) right? thats how you got that parabolas shape?

  20. lilsis76
    • 2 years ago
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    im sorry im kinda new to this lol

  21. lilsis76
    • 2 years ago
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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?

  22. Aperogalics
    • 2 years ago
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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:)

  23. Aperogalics
    • 2 years ago
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    @lilsis76

  24. lilsis76
    • 2 years ago
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    okay. im trying to do PF, is that ...getting the 2*1/4?

  25. Aperogalics
    • 2 years ago
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    PF??????????

  26. lilsis76
    • 2 years ago
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    hmm... let me try again

  27. lilsis76
    • 2 years ago
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    1/4th?

  28. lilsis76
    • 2 years ago
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    WAIT let me try again

  29. lilsis76
    • 2 years ago
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    i got..... 4 - 1/4 / 2-0

  30. lilsis76
    • 2 years ago
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    |dw:1352587321957:dw| i dont know how to work out the fraction and whole number

  31. Aperogalics
    • 2 years ago
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    slope m=|dw:1352587256238:dw| solve for x i just compare the two slopes:)

  32. lilsis76
    • 2 years ago
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    15/4/2--> 15/8

  33. lilsis76
    • 2 years ago
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    @Aperogalics

  34. Aperogalics
    • 2 years ago
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    @lilsis76 is it value of x?????????

  35. lilsis76
    • 2 years ago
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    yes? how do i know?

  36. Aperogalics
    • 2 years ago
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    i have to solve:(

  37. lilsis76
    • 2 years ago
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    oh. well then imma say yes it is then

  38. Aperogalics
    • 2 years ago
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    |dw:1352587871193:dw|

  39. lilsis76
    • 2 years ago
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    how did you get the 13/8?

  40. Aperogalics
    • 2 years ago
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    |dw:1352588075547:dw| so x=-1/8 or 2

  41. lilsis76
    • 2 years ago
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    haha okay. thanks

  42. lilsis76
    • 2 years ago
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    ugh factoring

  43. Aperogalics
    • 2 years ago
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    coordinates are (x,x^2) from fig x<0 x cannot be =2 Q=(-1/8,1/16)

  44. lilsis76
    • 2 years ago
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    okay i got the -1/8

  45. lilsis76
    • 2 years ago
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    and the 2, did you just add the 1/8 th twice to get 1/16?

  46. Aperogalics
    • 2 years ago
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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|

  47. Aperogalics
    • 2 years ago
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    no @lilsis76 as the coordinates are(x,x^2)

  48. Aperogalics
    • 2 years ago
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    @lilsis76 did u got it?????????

  49. lilsis76
    • 2 years ago
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    kinda, not really. okay well i ended up getting the - 1/8 for x, i plug it in right?

  50. Aperogalics
    • 2 years ago
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    yup:)

  51. lilsis76
    • 2 years ago
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    ok i got 8x+1 an x-2, x= -1/8, x=2

  52. Aperogalics
    • 2 years ago
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    now go ahead for shortcut:)

  53. lilsis76
    • 2 years ago
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    the shortcut u gave me?

  54. Aperogalics
    • 2 years ago
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    yup:)

  55. lilsis76
    • 2 years ago
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    i got -1/8

  56. lilsis76
    • 2 years ago
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    @Aperogalics

  57. lilsis76
    • 2 years ago
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    anywho, i did the math cuz shortcut got me lost. i have x = -1/8 and x=2

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