Find points $A$ and $B$ on the parabola $y=1-x^2$ such that an equilateral triangle is formed by the $x$-axis and the tangents $A$ and $B$.

Question

Question

Using Calculus, find points $A$ and $B$ on the parabola $y=1-x^2$ such that an equilateral triangle is formed by the $x$-axis and the tangents $A$ and $B$.

If I'm not mistaken, I think it would look roughly something like this.

Answer 1

HINT

The tangent line at point $x_0$ is given by: $$xf'(x_0)+b=0$$ or $$x\tan \alpha + b = 0 $$ where $\alpha$ is an angle between the line and x-axis.

Answer 2

Using your diagram, the $x$-intercepts would be $(-c,0)$ and $(c, 0)$, the $y$-intercept would be $(0,c\sqrt3)$ (Pythagorean theorem). We don't care what $c$ is.

$y'=-2x = \frac {a\sqrt3-0} {0-a} =-\sqrt3$ (slope of hypotenuse of right triangle on the right), so $x = \frac{\sqrt3} 2$.

then $y = 1 - (\frac{\sqrt 3} 2)^2 = \frac 1 4$.

So: $B(\frac{\sqrt 3} 2, \frac 1 4)$. By symmetry or using the same process, $A(-\frac{\sqrt 3} 2, \frac 1 4)$.