Thread: 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.

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)