geometrylinear algebralocus
I don't understand how to go on with this question:
"As the point $R$ moves on the line $x+y=1$ from $(1,0)$ to $(0,1)$, the point $P$ moves such that it has the same $x$-coordinate as $R$, and its $y$-coordinate is equal to the square root of that of $R$. Describe the locus of $P$ and draw the loci of both $R$ and $P$ on the same set of axes."
Any help would be appreciated,
Thanks
$$\begin{align*}3^2\left[(x-1)^2+(y-3)^2\right] =& (x-8)^2 + (y-2)^2\\ 9\left(x^2-2x+y^2-6y+10\right) =& x^2-16x+y^2-4x+68\\ 8x^2-2x+8y^2-50y+22 =& 0\\ x^2-\frac14x+y^2-\frac{25}4y+\frac{11}4=&0\\ \left(x-\frac{1}{8}\right)^2+\left(y-\frac{25}8\right)^2=&\frac{225}{32}\\ \left(x-\frac{1}{8}\right)^2+\left(y-\frac{25}8\right)^2=&\left(\frac{15\sqrt2}{8}\right)^2 \end{align*}$$
Let the center be at $(\theta,1)$. To reach this position, the circle must have rolled by the angle $\theta$, clockwise. Hence the absolute position of $P$ is the position of the center plus the position of $P$ relative to the center,
$$(x,y)=(\theta,1)+\left(-\sin\theta,-\cos\theta\right)=\left(\theta-\sin\theta,1-\cos\theta\right).$$
You can eliminate $\theta$ using
$$\cos\theta=1-y$$ and
$$x=\pm\arccos(1-y)\mp\sqrt{1-(1-y)^2}+2k\pi.$$
Best Answer
First, just graph the line first on the Cartesian plane. And then, determine the relationship between $R$ and $P$. Any point $R$ would be of the form $\left(t,1-t\right)$, where $0\leq t\leq1$.
This means that $P$ will have coordinates $\left(t,\sqrt{1-t}\right)$. So actually what you have is \begin{eqnarray*} f:R & \rightarrow & P\\ \left(t,1-t\right) & \mapsto & \left(t,\sqrt{1-t}\right). \end{eqnarray*} After some changes in the coordinates, what you have is $\left(1-s,s\right)\mapsto\left(1-s,\sqrt{s}\right)$, where $0\leq s\leq1$. Just run through all the points and you'll eventually see a graph, which is actually $y=\sqrt{1-x}$ on $\left[0,1\right]$.