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Here's the question- Find the maximum area of an isosceles triangle inscribed in the ellipse $x^2/a^2 + y^2/b^2 = 1$. My teacher solved it by considering two arbitrary points on the ellipse to be vertices of the triangle, being $(a\cos\theta, b\sin \theta)$ and $(a\cos\theta, -b\sin \theta)$. (Let's just say $\theta$ is theta) and then proceeded with the derivative tests(which i understood) But, he didn't indicate what our $\theta$ was,and declared that these points always lie on an ellipse. Why so? And even if they do, what's the guarantee that points of such a form will be our required vertices? One more thing, I'd appreciate it if you could suggest another way of solving this problem. Thank you!

Krosin
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Ambica Govind
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  • By symmetry, it makes sense that the height of the triangle be aligned with one of the axes. But why not also try $(a \cos \theta, b \sin \theta)$ and $(-a \cos \theta, b \sin \theta)$? – mjw Oct 15 '20 at 15:37
  • see parametric equation of ellipse: – sirous Oct 15 '20 at 15:37

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If we believe the symmetry argument, it is a simple matter to optimize area:

$$A_1 = a b \sin \theta (1+ \cos \theta)$$

$$A_2 = a b (1+ \sin \theta) \cos \theta$$

$$0=\frac{\partial A_1}{\partial \theta}=a b [\cos\theta (1+ \cos \theta)-\sin^2 \theta] = ab [\cos^2 \theta - \sin^2 \theta + \cos \theta]$$

$$0=\frac{\partial A_2}{\partial \theta}=a b [\cos^2 \theta - \sin \theta(1+ \sin \theta)]= ab[\cos^2 \theta - \sin^2 \theta-\sin \theta]$$

Solutions to the first equation include a minimum at $\theta=\pi$ and maxima at $\theta = \mp \frac{\pi}{3}$.

Solutions to the second equation include a minimum at $\theta=-\frac{\pi}{2}$ and maxima at $\theta = \frac{\pi}{6}$ and $\frac{5\pi}{6}.$

Interesting, we get solutions aligned with the $x-$axis and the $y-$axis!

You can certainly solve the problem taking three arbitrary points on the ellipse, constraining them to lie on iscoceles triangles, and then use Lagrange multipliers. The solutions will turn out to be those found above.

mjw
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  • Thanks so much! Btw, what are Lagrange multipliers? I'm just a high school student:( – Ambica Govind Oct 17 '20 at 05:49
  • The method of Lagrange multipliers is often applicable for constrained optimization problems. There is a function to optimize (minimize or maximize) and also constraints that need to be satisifed. Here is a link to a page on Lagrange multipliers that may seem rather terse, but there are embedded links to definitions of extrema and gradient: https://mathworld.wolfram.com/LagrangeMultiplier.html – mjw Oct 18 '20 at 03:34
  • Here is an example of the method: https://math.stackexchange.com/a/3585570/655367 – mjw Oct 18 '20 at 03:37
  • One more example: https://math.stackexchange.com/a/3786612/655367 – mjw Oct 18 '20 at 03:43