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anonymous

  • 5 years ago

If (2,-8) is the coordinates of an end point of a focal chord of y^2=4ax, then the co-ordinates of the other end are?

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  1. amistre64
    • 5 years ago
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    the other end of the locus mmbblbbmmb .... yeah

  2. amistre64
    • 5 years ago
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    this is a parabola; opens to the left; centered at origin right

  3. amistre64
    • 5 years ago
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    (2,8)

  4. anonymous
    • 5 years ago
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    no you are confusing it with latus rectum. thats not what it is... its a line joinin the point (2,-8) to the other end of the parabola, PASSING through the focus.

  5. amistre64
    • 5 years ago
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    ooohhhh; so its like a degenerate latus rectum :)

  6. anonymous
    • 5 years ago
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    well you can say latus rectum is one kind of a focal chord

  7. amistre64
    • 5 years ago
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    We would have to plug in the coordinates to calibrate the parabola maybe? -8^2 = 4a(2) 64 = 8a a = 8; the focal point is at 0,8

  8. amistre64
    • 5 years ago
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    err... (8,0)

  9. amistre64
    • 5 years ago
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    they say rise of run; which is y/x...but then they stick x in front to confuse the issue lol

  10. amistre64
    • 5 years ago
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    lets create a line from (2,-8 to 8,0 and see where these meet)

  11. amistre64
    • 5 years ago
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    2,-8 -8,0 ---- -6,8 ; slope = -4/3

  12. amistre64
    • 5 years ago
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    slope = 4/3....

  13. amistre64
    • 5 years ago
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    0 = (4/3)(8) + b y = (4/3)x -32/3 y = sqrt(32x)

  14. amistre64
    • 5 years ago
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    32x = 16x^2 -256x +1024 0 = 16x^2 -224x + 1024; right?

  15. anonymous
    • 5 years ago
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    the parametric point of a parabola is (at^2,2at) to suit the point (2,-8) we have, (8.(-1/2)^2, 2.8.(-1/2)) thus t=-1/2 lets call this t1, and the point at the other end of the parabola is \[(at _{2} ^{2},2at _{2}^{})\] t1*t2=-1 so t2=2 so the other point is (8.2^2,2.8.2) which is (32,32)

  16. amistre64
    • 5 years ago
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    http://www3.wolframalpha.com/input/?i=y^2+%3D+32x+and+y+%3D+4x%2F3+-+32%2F3

  17. amistre64
    • 5 years ago
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    yes; 32,32 :)

  18. amistre64
    • 5 years ago
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    apparently i had the line right; but mighta got messed up in solving the system i created :)

  19. amistre64
    • 5 years ago
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    y^2 = 32x 3y = 4x - 32 (-8) y^2 -24y = 256 y^2 - 24y - 256 = 0 ; thats the one lol

  20. amistre64
    • 5 years ago
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    y = -8, y = 32 3(32) + 32 ----------- = x 4 3(8) + 8 = x 24 + 8 = x = 32

  21. amistre64
    • 5 years ago
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    your way loks faster tho...

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