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.

  • $\begingroup$ That seems right: $(0,2^{-1/2})$. The maximum angle for a given ellipse occurs at the endpoints of the minor axis. $\endgroup$ – rogerl Nov 29 '17 at 21:12
  • $\begingroup$ 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. $\endgroup$ – 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

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

  • $\begingroup$ The answer doesn't match, do you have any alternate solution it? $\endgroup$ – Jasmine Dec 3 '17 at 13:07
  • $\begingroup$ @Jasmine: The answer doesn't match what? $\endgroup$ – Blue Dec 3 '17 at 14:19
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    $\begingroup$ @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. $\endgroup$ – MalayTheDynamo Dec 3 '17 at 16:45
  • $\begingroup$ 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? $\endgroup$ – Jasmine Dec 3 '17 at 17:04
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    $\begingroup$ @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.) $\endgroup$ – Blue Dec 3 '17 at 23:10

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