The abscissa of the point on the curve 9y2 = x3 , the normal at which cuts off equal intercepts on the coordinate axes is

The abscissa of the point on the curve 9y2 = x3 , the normal at which cuts off equal intercepts on the coordinate axes is

  1. A

    2

  2. B

    4

  3. C

    -4

  4. D

    -2

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    Solution:

    Differentiating 9y2=x3 we have dydx=x26y

    Any point on the curve is of the form t2,t3/3 and so dydx at this point is t/2 Thus an equation of normal is yt3/3=(2/t)Xt2  This will intersect coordinate axes at 0,2t+t3/3 and t46+t2,0

    Hence we must have  2t+t33=t2+t462+t23=t+t36

    Clearly t = 2 satisfies, the last equation. Hence the abscissa of the required point is 4. (For t = 0, the normal meets both the axes only at origin.)

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