Question

If the coordinates of a point be given by the equation $x=a\left( 1-\cos \theta \right)$, \[y=a\sin \theta\], then the locus of the point will be
(a) A straight line
(b) A circle
(c) A parabola
(d) An ellipse

Answer 1

Hint: The locus of points with coordinates $x=a\left( 1-\cos \theta \right)$, \[y=a\sin \theta\] can be obtained by taking the square of both the equations.

Complete step-by-step solution -

The coordinates of the point are given as,
$x=a\left( 1-\cos \theta \right)$ and \[y=a\sin \theta \]
We have to form an equation in terms of $x$ and $y$. Let us name the given equations as below,
$x=a\left( 1-\cos \theta \right)\ldots \ldots \ldots (i)$
$y=a\sin \theta \ldots \ldots \ldots (ii)$
Now, we have to take the square of the equations. Considering equation \[(i)\] first, we get
$\begin{align}
  & {{x}^{2}}={{\left[ a\left( 1-\cos \theta \right) \right]}^{2}} \\
 & \Rightarrow {{x}^{2}}={{a}^{2}}{{\left( 1-\cos \theta \right)}^{2}} \\
\end{align}$
Since we know that \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\], we get
$\begin{align}
  & \Rightarrow {{x}^{2}}={{a}^{2}}\left( 1+{{\cos }^{2}}\theta -2\cos \theta \right) \\
 & \Rightarrow {{x}^{2}}={{a}^{2}}+{{a}^{2}}{{\cos }^{2}}\theta -2{{a}^{2}}\cos \theta \ldots \ldots \ldots (iii) \\
\end{align}$
Taking the square of equation \[(ii)\], we get
${{y}^{2}}={{a}^{2}}{{\sin }^{2}}\theta \ldots \ldots \ldots (iv)$
Now, we can add equations \[(iii)\] and \[(iv)\].
${{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{a}^{2}}{{\cos }^{2}}\theta -2{{a}^{2}}\cos \theta +{{a}^{2}}{{\sin }^{2}}\theta $
Clubbing terms together, we get
$\begin{align}
  & {{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{a}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta -2{{a}^{2}}\cos \theta \\
 & {{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{a}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)-2{{a}^{2}}\cos \theta \\
\end{align}$
Since we know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\], we can apply the same in above equation and simplify as below,
\[\begin{align}
  & {{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{a}^{2}}-2{{a}^{2}}\cos \theta \\
 & {{x}^{2}}+{{y}^{2}}=2{{a}^{2}}-2{{a}^{2}}\cos \theta \\
 & {{x}^{2}}+{{y}^{2}}=2{{a}^{2}}\left( 1-\cos \theta \right) \\
\end{align}\]
Using the trigonometric relation, $1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2}$, we can simplify the above equation as,
\[\begin{align}
  & {{x}^{2}}+{{y}^{2}}=2{{a}^{2}}\left( 2{{\sin }^{2}}\dfrac{\theta }{2} \right) \\
 & {{x}^{2}}+{{y}^{2}}=4{{a}^{2}}{{\sin }^{2}}\dfrac{\theta }{2} \\
\end{align}\]
We can express the RHS as a square,
\[{{x}^{2}}+{{y}^{2}}={{\left( 2a\sin \dfrac{\theta }{2} \right)}^{2}}\]
We can compare the above equation with the general equation of a circle \[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\]. Hence, we get the locus as a circle.
Therefore, we get the correct answer as option (b).


Note: It is important to make sure that you form the right equation by using the right operation on the given points. Suppose that we consider the operation $\dfrac{x}{y}$, we will end up obtaining the result as \[\dfrac{x}{y}=\tan \dfrac{\theta }{2}\], which represents a line. If we consider a unit circle having a point $P\left( x,y \right)$ on its circumference, then we can write $\tan \theta