Chapter 1.4, Problem 47E

To determine

To find: The parameterization for the witch by expressing the coordinates of $P$ in terms of $t$.

Answer to Problem 47E

The parametric equations are $x=2\cot t$ and $y=2{\sin }^{2}t$ for the interval $0<t<\pi$.

Explanation of Solution

Given information: The given equalities are as follows:

(i) $x=AQ$

(ii) $y=2-AB\sin t$

(iii) $AB\cdot AO={\left(AQ\right)}^2$

The graph is given below:

  

Calculation:

It is given that the line $y=2$ is parallel to $x$ axis, $OQ$ is a line segment on $y$ axis and $OQ\perp QA$ . So, $\Delta OQA$ is a right angle triangle. It is noticed that the $\angle OAQ=t$ , because the line $y=2$ is parallel to $x$ axis. In this right angle triangle,

  $\begin{array}{c}\tan t=\frac{OQ}{AQ}\\ \tan t=\frac{2}{x}\\ x=\frac{2}{\tan t}\\ x=2\cot t\end{array}$

Now, by the equation $AB\cdot AO={\left(AQ\right)}^2$.

  $\begin{array}{c}AB=\frac{{\left(AQ\right)}^2}{AO}\\ =\left(\frac{AQ}{AO}\right)\left(AQ\right)\\ =\cos t\left(x\right)\\ =\left(\cos t\right)\left(2\cot t\right)\end{array}$

Simplify further.

  $\begin{array}{c}AB=\cos t\left(\frac{2\cos t}{\sin t}\right)\\ =\frac{2{\cos }^{2}t}{\sin t}\end{array}$

Use the equation $y=2-AB\sin t$.

  $\begin{array}{c}y=2-\left(\frac{2{\cos }^{2}t}{\sin t}\right)\sin t\\ =2-2{\cos }^{2}t\\ =2{\sin }^{2}t\end{array}$

Therefore, the parametric equations are $x=2\cot t$ and $y=2{\sin }^{2}t$ for the interval $0<t<\pi$.