Lorelei Patterson

Answered

2022-07-18

Isosceles triangle inscribed in an ellipse.

Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse $\frac{{x}^{2}}{{a}^{2}}}+{\displaystyle \frac{{y}^{2}}{{b}^{2}}}=1$ 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 $\begin{array}{}\text{(1)}& {A}^{2}=(x-a{)}^{2}{y}^{2}=(x-a{)}^{2}{b}^{2}(1-{\displaystyle \frac{{x}^{2}}{{a}^{2}}}).\end{array}$

Hence $\frac{dA}{dx}}=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(x-a{)}^{2}(x+{\displaystyle \frac{a}{2}})=0.$.

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 $A={\displaystyle \frac{\sqrt{3}ab}{4}}$.

The given answer is 3A.

What mistake I made ? Is it possible to find the answer to this question if the triangle was scalene?

Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse $\frac{{x}^{2}}{{a}^{2}}}+{\displaystyle \frac{{y}^{2}}{{b}^{2}}}=1$ 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 $\begin{array}{}\text{(1)}& {A}^{2}=(x-a{)}^{2}{y}^{2}=(x-a{)}^{2}{b}^{2}(1-{\displaystyle \frac{{x}^{2}}{{a}^{2}}}).\end{array}$

Hence $\frac{dA}{dx}}=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(x-a{)}^{2}(x+{\displaystyle \frac{a}{2}})=0.$.

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 $A={\displaystyle \frac{\sqrt{3}ab}{4}}$.

The given answer is 3A.

What mistake I made ? Is it possible to find the answer to this question if the triangle was scalene?

Answer & Explanation

Helena Howard

Expert

2022-07-19Added 12 answers

Step 1

Everything is correct until you use the value of x to calculate the area. You should have ${A}^{2}=(-\frac{3a}{2}{)}^{2}{b}^{2}(1-\frac{1}{4})$

Step 2

which will give you the correct answer $A=\frac{3\sqrt{3}}{4}ab$

Everything is correct until you use the value of x to calculate the area. You should have ${A}^{2}=(-\frac{3a}{2}{)}^{2}{b}^{2}(1-\frac{1}{4})$

Step 2

which will give you the correct answer $A=\frac{3\sqrt{3}}{4}ab$

Awainaideannagi

Expert

2022-07-20Added 5 answers

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 $\mathrm{\u25b3}ABC$ be the isosceles triangle with maximal area inscribed in our unit circle. Since orthogonal projections preserve area ratios, $\mathrm{\u25b3}ABC$ is the projection of the triangle with area $\mathcal{A}$. Note that since we wish to maximise the area, $\mathrm{\u25b3}ABC$ is simply an equilateral triangle with area $\frac{3\sqrt{3}}{4}$.

Step 2

Hence, by preservation of area ratios, $\begin{array}{rl}\frac{\mathcal{A}}{\text{Area of ellipse}}& =\frac{[ABC]}{\text{Area of circle}}\\ \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathcal{A}& =\frac{3\sqrt{3}}{4\pi}\cdot \pi ab\\ & =\frac{3ab\sqrt{3}}{4}\end{array}$ which is our answer.

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 $\mathrm{\u25b3}ABC$ be the isosceles triangle with maximal area inscribed in our unit circle. Since orthogonal projections preserve area ratios, $\mathrm{\u25b3}ABC$ is the projection of the triangle with area $\mathcal{A}$. Note that since we wish to maximise the area, $\mathrm{\u25b3}ABC$ is simply an equilateral triangle with area $\frac{3\sqrt{3}}{4}$.

Step 2

Hence, by preservation of area ratios, $\begin{array}{rl}\frac{\mathcal{A}}{\text{Area of ellipse}}& =\frac{[ABC]}{\text{Area of circle}}\\ \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathcal{A}& =\frac{3\sqrt{3}}{4\pi}\cdot \pi ab\\ & =\frac{3ab\sqrt{3}}{4}\end{array}$ which is our answer.

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