Answer
The point $P$ is $(-\frac{\sqrt{3}}{2},\frac{1}{4})$
The point $Q$ is $(\frac{\sqrt{3}}{2},\frac{1}{4})$
Work Step by Step
If the triangle $\triangle ABC$ is an equilateral triangle, then the angle $\angle ABC$ must be $60^{\circ}$.
We can find the slope of the line $BA$:
$m = tan ~60^{\circ}$
$m = \sqrt{3}$
The slope of the line $BA$ is $\sqrt{3}$ so the slope at point $P$ must also be $\sqrt{3}$
We can find the point $x$ on the parabola where the slope is $\sqrt{3}$:
$y = 1-x^2$
$y' = -2x = \sqrt{3}$
$x = -\frac{\sqrt{3}}{2}$
When $x = -\frac{\sqrt{3}}{2},$ then $y = 1-(-\frac{\sqrt{3}}{2})^2 = \frac{1}{4}$
Therefore, the point $P$ is $(-\frac{\sqrt{3}}{2},\frac{1}{4})$
By symmetry, the point $Q$ is $(\frac{\sqrt{3}}{2},\frac{1}{4})$
