# Right triangle on an ellipse, find the area

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Suppose that I have a right triangle $\triangle ABC$ where the median of $BC$ is located on the origin $O$. The triangle has the following properties:

• $\angle ABC = 90^{\circ}$,
• $B=(r\cos(\alpha), t\sin(\alpha))$,
• $A=(-\sqrt{r^2\cos^2(\alpha) + t^2\sin^2(\alpha)},0)$,
• $C=(\sqrt{r^2\cos^2(\alpha) + t^2\sin^2(\alpha)},0)$,

where $\alpha = \angle BOC$ and $r,t$ some arbitrary real numbers.

Now I am going to let Geogebra draw trace of the segments $\overline{AB}$ and $\overline{BC}$ while $\alpha$ changes from $0^\circ$ to $180^\circ$ for the following cases:

1. In this case $r=t=1$. Point $B$ changes location whereas the points $A$ and $C$ stay the same because

$$\sqrt{r^2\cos^2(\alpha) + t^2\sin^2(\alpha)}=\sqrt{r^2\cos^2(\alpha) + r^2\sin^2(\alpha)}=r$$

It can easily be verified that by interpolation the traces lead to a semi-circle of an area $\pi r^2/2$.

1. In this case $r=1$, $t=1.5$; as you can verify by the equations, the points $A,B,C, O$ change location. Particularly the point $B$ moves on an ellipse.

Finally, let's come to the problem:

I tried but couldn't calculate the area beneath the traces like the first case for the second case. How can I formulate the area? Any help is appreciated.

• Nice animation! (+1) How did you do it? – hypergeometric May 12 '16 at 13:24
• Thanks. I used Geogebra. It's a very nice tool but still I am not able to use all the features. – newzad May 12 '16 at 18:55

It is quite clear that the bell-shaped curve is formed by an arc of the ellipse of parametric equation $(r\cos\theta,t\sin\theta)$, traced by point $B$, and by an arc of the envelope of the family of lines $BC$ and $AB$.

In the following I'll assume $t\ge r$ and consider only the curve for $x\ge0$, as it is symmetrical around the $y$ axis. It is then enough to consider the envelope of lines $BC$.

The equations of lines $BC$ can be expressed as a function of $\alpha$ as $F(x,y,\alpha)=0$, where: $$F(x,y,\alpha)= y\left(r \cos\alpha-\sqrt{r^2 \cos ^2\alpha +t^2 \sin ^2\alpha}\right)- t\sin\alpha\left(x-\sqrt{r^2 \cos ^2\alpha +t^2 \sin ^2\alpha}\right).$$ The equation of the envelope can be found from: $$F(x,y,\alpha)={\partial\over\partial\alpha}F(x,y,\alpha)=0,$$ which give the parametric equations for the envelope: \begin{align} x_{env}=& \frac{-r^3 \cos ^4\alpha+t^2 \sin ^2\alpha \cos\alpha \sqrt{r^2 \cos^2\alpha+t^2 \sin ^2\alpha}} {r \left(r \cos\alpha-\sqrt{r^2 \cos ^2\alpha +t^2 \sin ^2\alpha}\right)}+\\ &+\frac{r^2 \cos ^3\alpha\sqrt{r^2 \cos ^2\alpha +t^2 \sin ^2\alpha} -r t^2 \sin ^4\alpha-2 r t^2 \sin ^2\alpha \cos^2\alpha} {r \left(r \cos\alpha-\sqrt{r^2 \cos ^2\alpha +t^2 \sin ^2\alpha}\right)},\\ y_{env}=&\frac{t \sin ^3\alpha \cos\alpha \left(r^2-t^2\right)} {r \left(r\cos\alpha-\sqrt{r^2 \cos^2\alpha+t^2 \sin ^2\alpha}\right)}.\\ \end{align}

I represented in the picture below the region (gray lines) swept by segment $BC$, for $r=1$ and $t=2.4$. The red curve is the ellipse, while the yellow curve is the envelope found above.

The envelope has a cusp, located at $\alpha=\alpha_0\approx\pi/4$. We are interested in its outer part, corresponding to $\alpha_0\le\alpha\le\pi/2$. This part of the envelope meets the ellipse at a point $P$: the corresponding value of $\alpha$ can be found from the equation $(x_{env}/r)^2+(y_{env}/t)^2=1$. This has only one solution $\bar\alpha$ in the interval $(\pi/4,\pi/2)$, given by: $$\begin{cases} \displaystyle\bar\alpha= \arccos{t\over\sqrt{3r^2+t^2}}, &\text{for r<t\le\sqrt3r;}\\ \displaystyle\bar\alpha= \arccos{r\over\sqrt{t^2-r^2}}, &\text{for t\ge\sqrt3r.}\\ \end{cases}$$

The corresponding values $\bar\theta$ for the parameter of the ellipse can be easily computed from: $\bar\theta=\arcsin(y_{env}(\bar\alpha)/t)$.

We can then represent half the area under the bell-shaped curve as an integral: $${1\over2}Area=\int_0^t y\,dx= \int_{\pi/2}^{\bar\theta}t\sin\theta{d\over d\theta}(r\cos\theta)\,d\theta+ \int_{\bar\alpha}^{\pi/2} y_{env}{dx_{env}\over d\alpha}\,d\alpha.$$

The first integral is easy to compute, but the second one is not and I suspect it cannot be expressed in elementary form. For $r=1$ and $t=1.5$ these integrals can be evaluated to $1.09117$ and $0.245874$, so that $Area=2.75247$.

• the outer yellow line is actually a curve. – newzad May 8 '16 at 14:29
• Yes, it is the curve whose equation I wrote above. It is not easy however to find an expression for the coordinates of the point where the two curves meet. – Aretino May 8 '16 at 14:34
• Thank you for your great work. – newzad May 8 '16 at 14:38
• I've expanded a bit my answer. – Aretino May 10 '16 at 22:40
• Do you mind to share how did you evaluate the integral? Maybe you computed. I would kindly ask you to share the code. If you are correct, and I suppose you are, I may contact you if you don't mind. – newzad May 11 '16 at 20:26