Hello, BarlowBarlow1!
You are given a parabola: .y .= .1  x²
You have to find points P and Q on the parabola so that the triangle ABC
formed by the xaxis and the tangent lines at P and Q is an equliateral triangle. Code:

A*
/\
/  \
/  \
/ * \
/ *  * \
/*  *\
P*  *Q
/  \
/*+*\
B  C
Let point P be (p, 1p²) .and Q be (q, 1q²)
We know that the angles of an equilateral triangle are all 60°.
. . .  . . .  . . .  . . . . . . . . . . . . . . . . ._
Hence, the slope of BA is: .tan(60°) .= .√3
We also know that the slope of a tangent is: .y' .= .2x
. . Hence, at point P(p, 1p²), the slope is: .2p
.  . .  . .  . . . . . . . . _ .  . . .  . . . . . . ._
So we have: . 2p .= .√3 . . → . . p .= .½√3
.  . .  . . . . . . . . . . . ._
By symmetry: .q .= .½√3
.  . .  . .  . .  . . . . ._ . . . . . . . . . . . _
The points are: .P(½√3, ¾) .and .Q(½√3, ¾)