Let y=f(x) represent a parabola with focus (-12, 0) and directrix y=-12.

Question

Let $y=f(x)$ represent a parabola with focus $\left(-\dfrac12, 0\right)$ and directrix $y=-\dfrac12$. Then
$$S = \left\{x\in\mathbb{R} : \tan^{-1}\left(\sqrt{f(x)}\right) + \sin^{-1}\left(\sqrt{f(x)+1}\right) = \frac{\pi}{2}\right\}:$$

Accepted Answer

The parabola with the given focus/directrix satisfies $\left(x+	frac12ight)^2 = y+	frac14$, i.e. $y = x^2+x = f(x)$.

The equation becomes $	an^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2+x+1} = \pi/2$, which requires $0\le x^2+x+1\le1$, i.e. $x^2+x\le0$; combined with $x^2+x\ge0$ (needed for the square root), $x^2+x=0 \Rightarrow x=0,-1$.

So $S$ contains exactly two elements.

Solutions · 1