Finding coordinates with differentiation

Question

Find points P and Q on the parabola $y = 1 - x^2$ so that the triangle ABC formed by the x-axis and the tangent lines at P and Q is an equilateral triangle

How do you solve this?

Answer 1

Hint. The base angles of an equilateral triangle are both $\frac{\pi}{3}$

The slope of a tangent to a curve is given by the value of the derivative at a point.

The slope of a line tells you the tangent of the angle that line makes with the horizontal.

Elaborating further, since OP asked:

Let the points of tangency on the curve be represented by $\pm x_0$ (by symmetry). Now (considering the left hand side of the symmetrical parabola) you know that $-2x_0 = \tan \frac{\pi}{3}$. Once you solve for $x_0$, simply find the coordinate $y_0$ on the curve corresponding to that. Your points $P$ and $Q$ will be given by $(-x_0, y_0)$ and $(x_0,y_0$) respectively.