# Parabola Tangent

• May 19th 2007, 04:42 PM
BarlowBarlow1
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.
• May 19th 2007, 06:32 PM
Soroban
Hello, BarlowBarlow1!

Quote:

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)