1. Parabola Tangent

You are given a parabola (1 - x^2). You have to find points (P and Q( on the parabola so that the triangle ABC formed by the x-axis and the tangent lines at P and Q is an equliateral triangle.

2. 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 x-axis and the tangent lines at P and Q is an equliateral triangle.
Code:
                      |
A*
/|\
/ | \
/  |  \
/   *   \
/ *  |  * \
/*    |    *\
P*      |      *Q
/       |       \
-----/*-------+-------*\----
B         |         C

Let point P be (p, 1-p²) .and Q be (q, 1-q²)

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, 1-p²), 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)