The respective lengths of the perpendiculars drawn from the origin to the tangent and to the normal of the astroid x^2/3 + y^2/3 = a^2/3 are denoted by F and G.

Determine all possible real a that satisfy this relationship:

4*F² + G² = a³ – 18.

The gradient of the astroid  X^2/3 + y^2/3 = a^2/3  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          x₀^1/3y  +  y₀^1/3x  =  x₀y₀^1/3 + y₀x₀^1/3

                                                            =  x₀^1/3 y₀^1/3(x₀^2/3  +  y₀^2/3)

                                                            =  x₀^1/3y₀^1/3a₀^2/3

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 = x₀^1/3y₀^1/3a₀^2/3/sqrt(x₀^2/3 + y₀^2/3)

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

                        =  (x₀^2/3 +y₀^2/3)³ – 3x₀^2/3y₀^2/3(x₀^2/3 + y₀^2/3)

                        =  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