Maximum area of a strange triangle

Let tangent to any point $P(acosθ, bsinθ)$ on the ellipse $\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1$ be drawn.

Equation of tangent is $\dfrac{xcosθ}{a} + \dfrac{ysinθ}{b} = 1$

$OQ = |\dfrac{1}{\sqrt{\frac{cos^{2}θ}{a^{2}} + \frac{sin^{2}θ}{b^{2}}}}|$

Note that $PQ$ is the distance of center to normal at $P$

Equation of normal is $\dfrac{ax}{cosθ} - \dfrac{by}{sinθ} = a^{2} - b^{2}$

$PQ = |\dfrac{a^{2} - b^{2}}{\sqrt{\frac{a^{2}}{cos^{2}θ} + \frac{b^{2}}{sin^{2}θ}}}|$

$\Delta POQ = \frac{1}{2}(OQ)(PQ)$

$\Delta POQ = \dfrac{ab(a^{2} - b^{2})}{2(a^{2}tanθ + b^{2}cotθ)}$

It is easy to maximize this expression.

$\Delta POQ_{max} = \dfrac{a^{2} - b^{2}}{4}$

Substitute the values to get $\Delta POQ_{max} = 20$

Interesting fact: $OQ$ will always have slope 1 , and $P$ will always be the point where its tangent has slope $-1$ .

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 ~ \Longleftrightarrow ~ b^2x^2+a^2y^2=a^2b^2$

Let $P=(a\cos\theta; b\sin\theta)$ , so that in polar coordinates, $P=(r\cos\phi; r\sin\phi)$ with $r^2=a^2\cos^2\theta+b^2\sin^2\theta$ and $\tan\phi=\frac{b\sin\theta}{a\cos\theta}$ , so that $\boxed{~r^2=\dfrac{1+b^2\tan^2\theta}{1+\tan^2\theta}~}\text{ and   }\boxed{\tan\phi=\dfrac{b}{a}\tan\theta~}$

Slope of tangent is given by $y'$ . $b^2x^2+a^2y^2=a^2b^2\implies(b^2x^2+a^2y^2)'=(a^2b^2)'\Leftrightarrow 2b^2x+2a^2y·y'=0 \Leftrightarrow y'=\dfrac{-b^2x}{a^2y}$ .

Slope of perpendicular $OQ$ is $\dfrac{-1}{y'}=\dfrac{a^2y}{b^2x}=\dfrac{a^2b\sin\theta}{b^2a\cos\theta}$ , but this slope is $\tan\alpha$ : $\tan\alpha=\frac{a}{b}\tan\theta$ , and so we have: $\boxed{\tan\theta=\dfrac{b}{a}\tan\alpha~}\text{ , }\boxed{\tan\phi=\dfrac{b^2}{a^2}\tan\alpha~}\text{ and }\boxed{~r^2=\dfrac{a^4+b^4\tan^2\alpha}{a^2+b^2\tan^2\alpha}~}$

Now, $\tan\beta=\tan{\alpha-\phi}=\dfrac{\tan\alpha-\tan\phi}{1+\tan\alpha\tan\phi}$ : $\boxed{\tan\beta=\dfrac{(a^2-b^2)\tan\alpha}{a^2+b^2\tan^2\alpha}}$

$OQ=\frac{r}{\sqrt{1+\tan^2\beta}}$ and $PQ=\frac{r\tan\beta}{\sqrt{1+\tan^2\beta}}$ , so searched area is $\begin{aligned}A&=\frac{1}{2}OQ·PQ=\frac{1}{2}·\dfrac{r^2\tan\beta}{1+\tan^2\beta}=\frac{1}{2}·\dfrac{r^2(a^2-b^2)\tan\alpha~(a^2+b^2\tan^2\alpha)}{(a^2+b^2\tan^2\alpha)^2+(a^2-b^2)^2\tan^2\alpha}\\&=\frac{1}{2}·\dfrac{r^2(a^2-b^2)\tan\alpha~(a^2+b^2\tan^2\alpha)}{(a^4+b^4\tan^2\alpha)(1+\tan^2\alpha)}=\frac{1}{2}·\dfrac{(a^2-b^2)\tan\alpha}{1+\tan^2\alpha}\end{aligned}$

$\boxed{~A=\dfrac{a^2-b^2}{2}·\sin\alpha\cos\alpha~}$ $\boxed{\boxed{~A\text{ is maximal when }\alpha=\frac\pi4~}}$ $\boxed{\boxed{~A_{max}=\dfrac{a^2-b^2}{4}~}}$