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# AREA of ellipse inside triangle

Note by Dev Kothari
3 years, 7 months ago

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Suppose that the triangle has vertices $$(-1,0)$$, $$(1,0)$$ and $$(0,2)$$. The ellipse must be centred somewhere on the $$y$$-axis, and must pass through the origin, so must have equation $\frac{x^2}{a^2} + \frac{(y-b)^2}{b^2} \; = \; 1$ where $$a,b > 0$$ are the semimajor and semiminor axes (in some order). This ellipse must just touch the line $$2x+y=2$$. Substituting $$x=1-\tfrac12y$$ into the equation for the ellipse, we obtain a quadratic equation for $$y$$: $(4a^2+b^2)y^2 - 4(b^2 + 2a^2b)y + 4b^2 = 0$ This equation must have just one root, so must have zero discriminant. The discriminant is $16a^2b^2(a^2+b-1)$ so we deduce that $$b = 1-a^2$$. We can now proceed like Aleksey to maximise the area when $$a=\tfrac{1}{\sqrt{3}}$$. · 3 years, 6 months ago

Let's denote ellipse half-axes as $$a$$ and $$b$$ with $$b$$ being along the altitude. Let's now stretch the triangle along it's altitude $$\frac{a}{b}$$ times, then the inscribed ellipse will become inscribed circle of radius $$a$$ and the triangle will become isosceles triangle with a base of 2 and altitude of $$\frac{2a}{b}$$. The incircle radius of such triangle is easy to find using Heron's formula to be: $r=\sqrt{1-\frac{2}{\sqrt{1+\frac{4a^2}{b^2}}+1}}$ on the other hand, it's $$a$$. Equating the two after some algebra we get: $b=|1-a^2|$ Since $$a$$ for obvious reasons is smaller than half of the base, we can omit the absolute value sign. The area of the ellipse is: $s=\pi ab=\pi a(1-a^2)$ At maximum its derivative reaches zero: $s'=\pi(1-3a^2)$ Therefore $a=\frac{1}{\sqrt{3}},~b=\frac{2}{3},s=\frac{2\pi}{3\sqrt{3}}$ · 3 years, 7 months ago