Given the co-ordinates of only two points A (x1,y1) and B (x2,y2) on a graph where the amplitude is y2-y1, what is the equation for the sine curve?
The distance from A to B is half the period.
[Math] Finding the equation for a sine wave given only two points
algebra-precalculus
Related Solutions
At all points on the curve the distance from $P$ to $A$ should be three times the distance from $P$ to $B$.
I recommend first sketching this by hand. Then use the distance formula to find the equation of the curve. Finally, make sure both answers agree!
Let $d(P,Q)$ represent the distance between points $P$ and $Q$. The problem statement tells us that $$d(P,A) = 3 d(P,B).$$ Plug in $(x,y)$ for $P$ and the given values of $A$ and $B$ to find the equation for the curve.
Hint: After some algebra you will find it is one of the conic sections.
To me, it seems what you want is parametric equations for the projection onto the $y$–$z$ plane of the curve on the surface of revolution$rz=1$ whose projection onto the $x$–$y$ plane is some clockwise logarithmic spiral
$$\begin{align} \theta &=\tfrac{\pi}{2}t & r &= \phi^{-(at+b)}\text{.} \end{align}$$
If that's the case, then the equations must be of the form
$$\begin{align} y(t)&=\phi^{-(at+b)}\sin \tfrac{\pi t}{2}\text{,} & z(t)&=\phi^{at+b} \end{align}$$ up to reparametrization, e.g.,
$$\begin{align}
y(t)&=\phi^{-t/2}\sin \tfrac{\pi t}{2}\text{,} &
z(t)&=\phi^{t/2}\text{,} &
t&\in [0,10]\text{:}
\end{align}$$
Best Answer
Since $A$ is a minimum and $B$ is a maximum, there exists some $n\in\Bbb Z_{\ge0}$ such that $$2n+1=\frac{|x_2-x_1|}{\lambda/2}$$ where $\lambda$ is the length of the wave. So $$\lambda=\frac{2|x_2-x_1|}{2n+1}$$ Now we have to translate the function $(y_2-y_1)\cos(2\pi x/\lambda)$ to make it to pass trough $B$:
$$f(x)=\frac{y_2-y_1}2\cos\left(\frac{\pi(x-x_2)(2n+1)}{|x_2-x_1|}\right)+\frac{y_1+y_2}2$$
EDIT: Since you have just added the condition that the curve from $A$ to $B$ is a half of the period just put $n=0$.