Find the points P and Q on the graph of f(x) = 1 - X^2 so that the triangle formed by the tangents at P and Q and the x-axis is equilateral.
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let the point of contact of the tangent with the curve in quadrant I be P(a,b)
dy/dx = -2x
so at P slope of the tangent is -2a
to be an equilateral triangle, the tangent must form a 60° angle at the x-axis, that is, the slope of that tangent must be -√3
then -2a = -√3
a = √3/2
P can also be called (a,1-a^2)
so P is (√3/2, 1 - 3/4) or (√3/2, 1/4
similarly, the slope at Q is +√3
so Q is (-√3/2, 1/4)