The least positive vlaue of the parameter 'a' for which there exist atleast one line that is tangent to the graph of the curve `y= x ^(3)-ax,` at one point and normal to the graph at another point is `p/q,` where p and q ar relatively prime positive integers. Find product pq.

Options

A. `(-oo, -4//3] `
B. `[-4//3,oo) `
C. `[4//3,oo) `
D. `(-oo,4//3]`

Answer

C

Explanation

The coordinates of any point P on ` y=x^(3)-ax ` are `(t,t^(3)-at)` and the equation of the tangent at P is ` y=(3t^(2)-a)x-2t^(3) .`
The abscissae of the point of intersection of ` y=x^(3)-ax ` and the tangent `y=(3t^(2)-a)x-2t^(3)` are the roots of the equation
` (3t^(2)-a)x-2t^(3)=x^(3)-ax `
` rArr x^(3)-3t^(2)x+2t^(3)=0 rArr (x-t)^(2)(x+2t)=0 rArr x=-2t,t `
Clearly, `x=t` corrcsponds to point P. So, the tangent at P cuts the curve again at ` Q(-2t,-8t^(3)+2at ). `
Now,
` y=x^(3)-ax rArr (dy)/(dx) = 3x^(2)-a rArr ((dy)/(dx))_(Q)=12t^(2)-a `
So, the slope of the normal at Q is ` -(1)/((dy)/(dx))_(Q)=-(1)/(12t^(2)-a) `
It is given that the tangent at P is normal at Q.
` therefore ((dy)/(dx))_(P)= -(1)/(((dy)/(dx))_(Q)) rArr (3t^(2)-a)=-(1)/(12t^(2)-a) `
` rArr (3t^(2)-a)(12t^(2)-a)+1=0 `
` rArr 36t^(4)-15 at^(2) + (a^(2)+1)=0 `
` rArr 36 u^(2)-15a u+(a^(2)+1)=0, " where " t^(2)=u. `
This equation must have at least one real root u and ` t^(2)=u ` implies that the root must be positive.
` therefore (15a)/(72) gt 0 " and " 225a^(2)-144(a^(2)+1) ge 0 `
` [ "Using: " -(b)/(2a) gt 0 " and Disc " c ge 0 ] `
`rArr a gt 0 " and " 9(3a-4)(3a+4) ge 0 `
` rArr a ge (4)/(3) rArr a in [4//3, oo) `