# 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.

• That seems right: $(0,2^{-1/2})$. The maximum angle for a given ellipse occurs at the endpoints of the minor axis. – rogerl Nov 29 '17 at 21:12
• I would change the statement of the problem to say that the segment between its foci subtends a right angle from no point on the boundary. The distance between the foci is a number (as is the circumference), and numbers don't subtend angles. – John Hughes Dec 3 '17 at 13:14

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

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

• The answer doesn't match, do you have any alternate solution it? – Jasmine Dec 3 '17 at 13:07
• @Jasmine: The answer doesn't match what? – Blue Dec 3 '17 at 14:19
• @Blue The Given solution, possibly. This was probably a homework question, and OP couldn't find the answer. Now that she has got the answer, it turns out the gotten and the given do not match. – MalayTheDynamo Dec 3 '17 at 16:45
• No, the solution is not wrong. only difference is that 0 is not included in the range. And even it's true as 0 can't be eccentricity for an ellipse? – Jasmine Dec 3 '17 at 17:04
• @Jasmine: Just as a square is a rectangle with all sides equal, a circle is an ellipse with major and minor radii equal. From that "inclusive" point of view (which I generally believe is most-correct), $0$ is the eccentricity of an ellipse. That said, it is not unreasonable to exclude $0$ and restrict attention to non-circular ellipses, which the author of the problem may have done. (The exclusion seems unnecessary here, since the result (and the proof) is valid for circles, too.) – Blue Dec 3 '17 at 23:10