Find the range of eccentricity of an ellipse such that the distance between its foci doesn't subtend any right angle on its circumference. 
What is the range of eccentricity of ellipse such that its foci don't subtend any right angle on its circumference?

I thought that the eccentricity would definitely be more than $0$ and less than $\frac{1}{\sqrt2}$ The latter value is for an ellipse with $ae=b$, in which a right angle is subtended on an endpoint of the minor axis.
 A: Your answer is correct, except that $0$ should be included (there are no right angles subtended in a circle).
Here's a complete solution:

Using the parameterization $P=(a \cos\theta, b\sin\theta)$ for an origin-centered ellipse with major radius $a$ (in the $x$ direction) and minor radius $b$ (in the $y$ direction), consider the foci at points $F_{\pm}=(\pm c, 0)$, where $a^2 = b^2 + c^2$.
$\angle F_{+}PF_{-}$ will be a right angle if and only if 

$$(F_{+}-P)\cdot(F_{-}-P) = 0 \tag{$\star$}$$

That is,
$$\begin{align}
0 &= (c - a \cos\theta )(-c-a\cos\theta) + (0 - b \sin\theta)(0-b\sin\theta) \\[4pt]
&= -c^2 + a^2 \cos^2\theta + b^2\sin^2\theta \\[4pt]
&= -c^2+a^2\cos^2\theta + ( a^2-c^2)(1-\cos^2\theta) \\[4pt]
&= a^2 - 2 c^2 + c^2 \cos^2\theta \tag{1}
\end{align}$$
Writing $c = ae$, where $e$ is the eccentricity, we can factor-out $a^2$ to get
$$e^2\cos^2\theta = 2 e^2 - 1 \tag{2}$$
In order for $(2)$ to be solvable for $\theta$, we obviously must have $e\neq 0$ (so that $\theta$ appears in the equation at all); then, for non-zero $e$, since $0\leq \cos^2\theta \leq 1$, the solvability of $(2)$ requires
$$
0 \leq 2-\frac{1}{e^2}\leq 1 \quad\to\quad 
2 \geq \frac{1}{e^2}\geq 1 \quad\to\quad
\sqrt{\frac{1}{2}} \leq e \leq 1 \tag{3}$$
In other words, the equation is not solvable for $\theta$ ---that is, there are no subtended right angles--- for $e < {1\over\sqrt{2}}$ or $e > 1$ (although we dismiss the latter possibility, as such eccentricities belong to hyperbolas). Therefore, the desired range of eccentricities is

$$0 \leq e < \frac{1}{\sqrt{2}} \tag{$\star\star$}$$

