If the tangent to the curve 2y3=ax2+x3 at the point (a, a) cuts off intercepts α and β on the coordinate axes such that α2+β2=61, then a=

If the tangent to the curve 2y3=ax2+x3 at the point (a, a) cuts off intercepts α and β on the coordinate axes such that α2+β2=61, then a=

  1. A

    ± 30

  2. B

    ± 5

  3. C

    ± 6 

  4. D

    ± 61 

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    Solution:

    We have,

                               2y3=ax2+x3 6y2dydx=2ax+3x2dydx=2ax+3x26y2dydx(a,a)=56

    The equation of the tangent at (a, a) is

    ya=56(xa)5x6y+a=0

    This intercepts lengths a/5 and a/6 with x and y-axes respectively.

     α=a/5 and β=a/6

    Now,

    α2+β2=61a225+a236=61a2=25×36a=±30

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