 Lorelei Patterson

2022-07-18

Isosceles triangle inscribed in an ellipse.
Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$ with its (unique) vertex lying at one of the ends of the major axis of the ellipse.
The three vertices of the triangle would be (a,0), (x,y), (x,-y) .
The area of the triangle by Heron's formula is $\begin{array}{}\text{(1)}& {A}^{2}=\left(x-a{\right)}^{2}{y}^{2}=\left(x-a{\right)}^{2}{b}^{2}\left(1-\frac{{x}^{2}}{{a}^{2}}\right).\end{array}$
Hence $\frac{dA}{dx}=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left(x-a{\right)}^{2}\left(x+\frac{a}{2}\right)=0.$.
We have minimum at $x=a$ and maximum at $x=-\frac{a}{2}$.
Substituting back in (1) and taking square roots on both the sides gives $A=\frac{\sqrt{3}ab}{4}$.
What mistake I made ? Is it possible to find the answer to this question if the triangle was scalene? Helena Howard

Expert

Step 1
Everything is correct until you use the value of x to calculate the area. You should have ${A}^{2}=\left(-\frac{3a}{2}{\right)}^{2}{b}^{2}\left(1-\frac{1}{4}\right)$
Step 2
which will give you the correct answer $A=\frac{3\sqrt{3}}{4}ab$ Awainaideannagi

Expert

Step 1
We perform an orthogonal projection to map the ellipse to the unit circle.
Let the maximal area of our isosceles triangle be $\mathcal{A}$ which we wish to find, and let $\mathrm{△}ABC$ be the isosceles triangle with maximal area inscribed in our unit circle. Since orthogonal projections preserve area ratios, $\mathrm{△}ABC$ is the projection of the triangle with area $\mathcal{A}$. Note that since we wish to maximise the area, $\mathrm{△}ABC$ is simply an equilateral triangle with area $\frac{3\sqrt{3}}{4}$.
Step 2
Hence, by preservation of area ratios, which is our answer.

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