Miscellaneous · Q5: Largest isosceles triangle in an ellipse, apex at $(a,0)$⁠: area $\frac{3\sqrt3}{4}ab$

Basics you need

Solution

1. The base must be a vertical chord

   Let the other two corners be $P$ and $Q$⁠, with $AP = AQ$⁠. For a point $(x, y)$ of the ellipse, Basic 2 gives its squared distance from $A(a, 0)$ as a function of $x$ alone:

   $$\begin{aligned} D(x) &= (x - a)^2 + y^2 \\ &= (x - a)^2 + \tfrac{b^2}{a^2}(a^2 - x^2). \end{aligned}$$

   Its derivative is

   $$\begin{aligned} D'(x) &= 2(x - a) - \tfrac{2b^2}{a^2}x \\ &= -2\left[a - x\left(1 - \tfrac{b^2}{a^2}\right)\right]. \end{aligned}$$

   Since $a \ge b > 0$⁠, we have $0 \le 1 - \frac{b^2}{a^2} < 1$⁠. So if $x \le 0$ the product $x\left(1 - \frac{b^2}{a^2}\right)$ is $\le 0 < a$⁠; if $0 < x \le a$ it is less than $x \le a$⁠. Either way the bracket is positive, so $D'(x) < 0$ on $[-a, a]$ and $D$ is strictly decreasing (Theorem 1). Hence $AP = AQ$ forces $P$ and $Q$ to have the same $x$⁠, and then their $y$-values are $\pm y$⁠:

   $$P(x, y),\quad Q(x, -y),\quad y = \tfrac{b}{a}\sqrt{a^2 - x^2}.$$
   The question does not tell us where the base is, so we must prove it. As $x$ runs from $a$ down to $-a$⁠, the distance from $A$ to the point of the ellipse above (or below) $x$ keeps growing, so it never takes the same value at two different $x$⁠. Two points equally far from $A$ must therefore share their $x$⁠, one above and one below the axis.
   Points of the ellipse (Basic 2); Theorem 1, Section 6.3 (Basic 4).

2. The area as a function of $x$⁠, and its range

   The base $PQ$ is vertical with length $2y = \frac{2b}{a}\sqrt{a^2 - x^2}$⁠. The height is the horizontal distance from $A(a, 0)$ to the line through $P$ and $Q$⁠, which is $a - x$⁠. So

   $$\begin{aligned} A(x) &= \tfrac12 \cdot \tfrac{2b}{a}\sqrt{a^2 - x^2}\cdot(a - x) \\ &= \tfrac{b}{a}(a - x)\sqrt{a^2 - x^2}. \end{aligned}$$

   Range. We need $y > 0$ (otherwise $P = Q$ and there is no triangle), that is $-a < x < a$⁠. At both ends the triangle collapses and $A \to 0$⁠.

   For a vertical base, the perpendicular from the apex is horizontal, so the height is just the difference of $x$-coordinates, $a - x$⁠, which is positive because $x < a$⁠.
   Area $= \frac12$ base $\times$ height. Basics: building the function to optimise

3. Differentiate and factorise

   Product rule with $u = a - x$⁠, $u' = -1$⁠, $v = \sqrt{a^2 - x^2}$⁠, $v' = \frac{-x}{\sqrt{a^2 - x^2}}$⁠:

   $$\begin{aligned} A'(x) &= \tfrac{b}{a}\left[-\sqrt{a^2 - x^2} - \tfrac{x(a - x)}{\sqrt{a^2 - x^2}}\right] \\ &= \tfrac{b}{a}\cdot\tfrac{-(a^2 - x^2) - x(a - x)}{\sqrt{a^2 - x^2}} \\ &= \tfrac{b}{a}\cdot\tfrac{2x^2 - ax - a^2}{\sqrt{a^2 - x^2}} \\ &= \tfrac{b}{a}\cdot\tfrac{(2x + a)(x - a)}{\sqrt{a^2 - x^2}}. \end{aligned}$$
   Putting both terms over $\sqrt{a^2 - x^2}$ gives one fraction whose numerator is a quadratic. Expand: $-a^2 + x^2 - ax + x^2 = 2x^2 - ax - a^2$⁠. Check the factorisation: $(2x + a)(x - a)$ is $2x^2 - 2ax + ax - a^2$⁠, which is $2x^2 - ax - a^2$⁠.
   Product rule (Section 5.3, rule (2)) and chain rule (Theorem 4, Section 5.3.1) (Basic 3).

4. Sign of $A'$ on the whole range

   For $-a < x < a$⁠: $\frac ba > 0$⁠, $\sqrt{a^2 - x^2} > 0$ and $x - a < 0$⁠. So $A'(x)$ has the opposite sign to $2x + a$⁠:

   • for $-a < x < -\frac a2$⁠: $2x + a < 0$⁠, so $A'(x) > 0$⁠;
   • at $x = -\frac a2$⁠: $A'(x) = 0$⁠;
   • for $-\frac a2 < x < a$⁠: $2x + a > 0$⁠, so $A'(x) < 0$⁠.

   The only critical point in the range is $x = -\frac a2$ (the other zero, $x = a$⁠, is excluded).

   Of the four factors, three have a fixed sign on the whole range, so only $2x + a$ decides. This gives the sign everywhere at once, which is exactly what a global conclusion needs.
   Basics: sign of a product

5. The greatest area, on the whole range

   By Theorem 1, $A$ is increasing on $\left(-a, -\frac a2\right]$ and decreasing on $\left[-\frac a2, a\right)$ (strictly, by its proof). So $A(x) < A\!\left(-\frac a2\right)$ for every other $x$ in the range: $x = -\frac a2$ gives the greatest area, not just a local one. Now compute it:

   $$\begin{aligned} a - x &= \tfrac{3a}{2}, \quad \sqrt{a^2 - \tfrac{a^2}{4}} = \tfrac{\sqrt3}{2}a, \\ A\!\left(-\tfrac a2\right) &= \tfrac ba \cdot \tfrac{3a}{2}\cdot\tfrac{\sqrt3\,a}{2} = \tfrac{3\sqrt3}{4}ab. \end{aligned}$$

   The corners of this triangle are $A(a, 0)$ and $\left(-\frac a2, \pm\frac{\sqrt3}{2}b\right)$⁠: its base is $\sqrt3\,b$ and its height $\frac{3a}{2}$⁠.

   The first derivative test (Theorem 3) would only say "local maximum". Using the sign of $A'$ on the entire interval $(-a, a)$ compares $x = -\frac a2$ with every other allowed triangle, including the thin ones near the ends where $A \to 0$⁠.
   Theorem 1, Section 6.3 (Basic 4). Lesson: optimisation problems

6. The other cases give the same answer

   The other end $(-a, 0)$⁠. The reflection $(x, y) \mapsto (-x, y)$ maps the ellipse onto itself (its equation contains only $x^2$ and $y^2$⁠) and keeps lengths and areas. It turns each triangle with its vertex at $(a, 0)$ into one with its vertex at $(-a, 0)$⁠, and back. So the largest area is again $\frac{3\sqrt3}{4}ab$⁠.

   The case $b > a$⁠. Now the major axis lies along the $y$-axis, with ends $(0, \pm b)$⁠. The reflection $(x, y) \mapsto (y, x)$ keeps areas and turns the ellipse into

   $$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1,$$

   whose major axis is on the $x$-axis, with ends $(\pm b, 0)$⁠. Steps 1 to 5 with $a$ and $b$ interchanged give the maximum $\frac{3\sqrt3}{4}ba$⁠, the same number.

   So in every case the maximum area is $\frac{3\sqrt3}{4}ab$.

   The answer is symmetric in $a$ and $b$⁠, which is why the labelling does not matter. Step 1 does use that the vertex is at an end of the major axis: $a \ge b$ is what makes $D'$ negative on the whole of $[-a, a]$⁠. (A further fact, not proved in this book: no triangle inscribed in the ellipse has area more than $\frac{3\sqrt3}{4}ab$⁠. So even a triangle with a corner at $A$ where the equal sides do not meet cannot do better.)

Every fact used, and where it comes from