How can we find the equation of the curve as shown in the figure?

Question

Above are the illustrations, below in the red box is the QUESTION '$a$' and '$b$' are the length of the lines as shown in the figure. The two lines are divided into '$n$' equal parts and lines are drawn according to the illustration. The curve formed by the intersection of those lines is the curve with blue stroke. $\theta = \text{Angle between the two lines}$

Answer 1

Given $O(0,0)$, $A(a\cos\theta,a\sin\theta)$, $B(b,0)$ consider points $P_t$, $P_{t+\Delta t}$ on $OA$, $Q_t$, $Q_{t+\Delta t}$ on $OB$:

\begin{align} P_t&=A\,(1-t) ,\\ Q_t&=B\,(t+\Delta t) ,\\ P_{t+\Delta t}&=A\,(1-(t+\Delta t)) ,\\ Q_{t+\Delta t}&=B(t+2\Delta t) . \end{align}

Lines $P_tQ_t$, $P_{t+\Delta t}Q_{t+\Delta t}$ intersect at a point $z=(x(t),y(t))$, \begin{align} x(t)&= (\Delta t+1)^{-1}(2b \Delta t^2 +(a\cos\theta (t-1)+3 t b)\Delta t \\ &+a\cos\theta-2ta\cos\theta+(b+a\cos\theta)t^2) ,\\ y(t)&=(\Delta t+1)^{-1} (t-1+\Delta t)(t-1)a\sin\theta . \end{align}

\begin{align} \lim_{\Delta t \to 0}x(t) &= {a\cos\theta-2t a\cos\theta+(b+a\cos\theta)t^2} \\ &= a\cos\theta\cdot(1-t)^2+0\cdot2(1-t)t+b\cdot t^2 ,\\ \lim_{\Delta t \to 0}y(t) &= {(t-1)^2 a\sin\theta} \\ &= a\sin\theta\cdot(t-1)^2+0\cdot2(1-t)t+0\cdot t^2 ,\\ \\ (x(t),\,y(t))&=A\cdot(1-t)^2+2 O\cdot(1-t)t+B\cdot t^2 , \end{align}

which is a well-known quadratic Bezier segment with control points $A,O,B$.

Answer 2

Here's a solution for $a=b=1$ and $\theta=\pi/2$. The result is trivially scaled for any $a$ and $b$, but generalizing for other angles might be more difficult.

First, let $d$ be our step size, so that:

$$d=\frac{1}{n}$$

Then according to the rules above we can see that:

$$x=kd \\ y=1+d-kd$$

Where $k=0,1,\dots,n$.

Let's find the equation of the $k$th straight line, which forms our curve. From two known points $(kd,0)$ and $(0,1+d-kd)$ we obtain the equation:

$$y_k=1+d-kd- \left( \frac{1+d}{kd}-1 \right) x$$

But the straight lines approach the curve only between the intersections with adjacent lines:

$$x_{k1}: \qquad y_{k-1}(x)=y_k(x) \\ x_{k2}: \qquad y_{k+1}(x)=y_k(x)$$

Solving the equations, we take the arithmetic mean as a point which belongs to both the straight line and the curve:

$$x_k=\frac{x_{k1}+x_{k2}}{2}=\frac{k^2 d^2}{1+d}$$

Thus, the equation for our curve at each point is described by:

$$y_k(x_k)=1+d-kd- \left( \frac{1+d}{kd}-1 \right) \frac{k^2 d^2}{1+d}$$

Or, after simplifications:

$$y_k(x_k)=1+d-2kd+\frac{k^2 d^2}{1+d}$$

Or:

$$y_k \left( \frac{k^2 d^2}{1+d} \right)=1+d-2kd+\frac{k^2 d^2}{1+d}$$

Let's introduce a new variable:

$$z=\frac{k^2 d^2}{1+d}$$

Then we have:

$$y \left( z \right)=1+d-2 \sqrt{(1+d)z}+z$$

Now we remember that our step size is supposed to be infinitely small, so we take the limit:

$$\lim_{d \to 0} y(z)=1-2 \sqrt{z}+z$$

Simplifying and renaming our variable we have:

$$y(x)=(1- \sqrt{x})^2$$

Our curve is supposed to be symmetric around $y=x$, which is true in this case.

Here's an illustration for $d=1/10$ compared to $y(x)$: