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 Getting Quadratic and Cubic with Astroid (Posted on 2011-10-22)
The respective lengths of the perpendiculars drawn from the origin to the tangent and to the normal of the astroid x2/3 + y2/3 = a2/3 are denoted by F and G.

Determine all possible real a that satisfy this relationship:

4*F2 + G2 = a3 – 18.

 See The Solution Submitted by K Sengupta No Rating

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 Solution | Comment 1 of 2
The gradient of the astroid  X2/3 + y2/3 = a2/3  is given by:   dy/dx = -(y/x)1/3

So a tangent at P(x0,y0) has the equation:   (y – y0)/(x – x0) = -(y0/x0)1/3

which simplifies to          x01/3y  +  y01/3x  =  x0y01/3 + y0x01/3

=  x01/3 y01/3(x02/3  +  y02/3)

=  x01/3y01/3a02/3

Now, using the fact that a line px + qy = c is at perpendicular distance of

c/sqrt(p2 + q2)  from the origin, we have  F = x01/3y01/3a02/3/sqrt(x02/3 + y02/3)

which simplifies to   F = (x0y0a)1/3.            (1)

Consider the triangle formed by the tangent at P, the line OP(= r) and the

perpendicular from O to the tangent, whose sides are G, F and r.

F2 + G2 =  r2      =  x02 + y02

=  (x02/3 +y02/3)3 – 3x02/3y02/3(x02/3 + y02/3)

=  a2  -  3(x0y0a)2/3

=  a2  -  3F2        using (1)

So  4F2 + G2 = a2 and the required relationship becomes  a2 = a3 – 18.

Thus  a3 – a2 – 18 = 0 giving:      (a – 3)(a2 +2a + 6) = 0,  whose only

real solution is  a = 3.

 Posted by Harry on 2011-10-24 23:31:45

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