Area of Shapes - 2

To prepare ourselves for polar integration, we want to choose $P$ in terms of radians. Let $P = (\sqrt{\cot(\theta)}, \sqrt{\tan(\theta)})$ . By doing this, $P$ is the point at an angle $+\theta$ radians from the positive $x$ -axis, located at the blue curve. (This can be done by substituting $y=\tan(\theta)x$ in the equation for the blue curve.)

Let now $t = \sqrt{\cot(\theta)}$ . The tangent to the curve at the point $P$ can be written as $y = \frac{t-x}{t^2} + \frac{1}{t} = \frac{2t-x}{t^2}$ . We want to find the perpendicular line to this tangent touching the curve in the fourth quadrant. We can do this, using the fact that the product of the slopes must be equal to $-1$ . So, the slope of the perpendicular line is equal to $t^2$ and so we can find that the perpendicular line has the form $y = t^2(x-\frac{1}{t})-t = t^2x-2t.$

Now, we will find the intersections of the two tangents. Notice that for the intersection $\frac{2t-x}{t^2} = t^2x-2t, \text{ and so } 2t - x = t^4x-2t^3, \text{ from which follows that } x = \frac{2t(t^2+1)}{t^4+1}.$ By plugging in this value for $x$ in one of the equations of the tangents, we find that $y = \frac{2t(t^2-1)}{t^4+1}$ .

We now want to find the area enclosed by the orange curve, which is, by symmetry and the polar integration equation, equal to $\mathcal{A} := 8 \cdot \frac{1}{2} \cdot \int_0^\frac{\pi}{4} r^2(t) \mathrm{d}\theta,$ where $r^2(t)$ is the squared length of the point $(x(t), y(t)) = \left(\frac{2t(t^2+1)}{t^4+1}, \frac{2t(t^2-1)}{t^4+1}\right)$ . Some calculations will lead to the conclusion that $r^2(t) = \frac{8t^2}{t^4+1}.$ Knowing that $t = \sqrt{\cot(\theta)}$ , we notice that $\mathcal{A} = 4 \cdot \int_0^\frac{\pi}{4} \frac{8|\cot(\theta)|}{\cot^2(\theta)+1} \mathrm{d}\theta = 32 \cdot \int_0^\frac{\pi}{4} \frac{\cot(\theta)}{\csc^2(\theta)} \mathrm{d}\theta = 32 \cdot \int_0^\frac{\pi}{4} \sin(\theta)\cos(\theta) \mathrm{d}\theta = 16 \int_0^\frac{\pi}{4} \sin(2\theta) \mathrm{d}\theta = -8\cos(2\theta) \bigg|_0^\frac{\pi}{4} = \boxed{8}.$

Edit: The shape of the collection of black dots looks like the following orange curve.