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Yahoo!

  • 2 years ago

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 ?

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  1. Yahoo!
    • 2 years ago
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    @shubhamsrg

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

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

  4. Yahoo!
    • 2 years ago
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    Nw wat to do ?

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

  6. shubhamsrg
    • 2 years ago
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    hence slope should be equal to 1 or -1.

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

  8. shubhamsrg
    • 2 years ago
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    |dw:1359206768852:dw|

  9. Yahoo!
    • 2 years ago
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    Did nt get u

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

  11. Yahoo!
    • 2 years ago
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    yup

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

  13. Yahoo!
    • 2 years ago
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    ok...then

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

  15. shubhamsrg
    • 2 years ago
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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.

  16. Yahoo!
    • 2 years ago
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    (a-0) / (0-a) = -1

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

  18. Yahoo!
    • 2 years ago
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    ok...

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

  20. Yahoo!
    • 2 years ago
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    yes

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

  22. Yahoo!
    • 2 years ago
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    Yeah got it..:) thxx

  23. shubhamsrg
    • 2 years ago
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    glad you did. ^_^

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