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Yahoo!

The x co-ordinate of a point on the curve 9y^2 = x^3 , the normal at which cuts off equal intercept on the axes is ?

  • one year ago
  • one year ago

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  1. Yahoo!
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    @shubhamsrg

    • one year ago
  2. Yahoo!
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    dy/dx =x^2/(6y)

    • one year ago
  3. Yahoo!
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    gradient of normal is -6y/(x^2)

    • one year ago
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    Nw wat to do ?

    • one year ago
  5. shubhamsrg
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    -6y/x^2 is the slope of a line which cuts of equal intercepts as x axis.

    • one year ago
  6. shubhamsrg
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    hence slope should be equal to 1 or -1.

    • one year ago
  7. shubhamsrg
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    "-6y/x^2 is the slope of a line which cuts of equal intercepts ON THE axes.

    • one year ago
  8. shubhamsrg
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    |dw:1359206768852:dw|

    • one year ago
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    Did nt get u

    • one year ago
  10. shubhamsrg
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    -6y/x^2 is the slope of the normal right ?

    • one year ago
  11. Yahoo!
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    yup

    • one year ago
  12. shubhamsrg
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    and we are given that the normal cuts of equal intercepts on the axes.

    • one year ago
  13. Yahoo!
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    ok...then

    • one year ago
  14. Yahoo!
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    let the Points be (a,0) and (0,a ) ryt ?

    • one year ago
  15. shubhamsrg
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    any line which cuts off equal intercepts on the axes must have a slope of 1 or -1 right? Since it is making an angle of pi/4 or 3pi/4 with the +ve X axes.

    • one year ago
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    (a-0) / (0-a) = -1

    • one year ago
  17. shubhamsrg
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    What I mean is -6y /x^2 = 1 and/or -1

    • one year ago
  18. Yahoo!
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    ok...

    • one year ago
  19. shubhamsrg
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    we also know y = x^(3/2) /3 or -x^(3/2) /3

    • one year ago
  20. Yahoo!
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    yes

    • one year ago
  21. shubhamsrg
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    substitute in the previous eqn to find the required value(s) of x

    • one year ago
  22. Yahoo!
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    Yeah got it..:) thxx

    • one year ago
  23. shubhamsrg
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    glad you did. ^_^

    • one year ago
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