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Question: If the coordinates of a point be given by the equation \(x=a\left( 1-\cos \theta \right)\), \[y=a\si...

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

Explanation

Solution

Hint: The locus of points with coordinates x=a(1−cos⁡θ)x=a\left( 1-\cos \theta \right), y=asin⁡θy=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(1−cos⁡θ)x=a\left( 1-\cos \theta \right) and y=asin⁡θy=a\sin \theta
We have to form an equation in terms of xx and yy. Let us name the given equations as below,
x=a(1−cos⁡θ)………(i)x=a\left( 1-\cos \theta \right)\ldots \ldots \ldots (i)
y=asin⁡θ………(ii)y=a\sin \theta \ldots \ldots \ldots (ii)
Now, we have to take the square of the equations. Considering equation (i)(i) first, we get
x2=[a(1−cos⁡θ)]2 ⇒x2=a2(1−cos⁡θ)2 \begin{aligned} & {{x}^{2}}={{\left[ a\left( 1-\cos \theta \right) \right]}^{2}} \\\ & \Rightarrow {{x}^{2}}={{a}^{2}}{{\left( 1-\cos \theta \right)}^{2}} \\\ \end{aligned}
Since we know that (a−b)2=a2−2ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}, we get
⇒x2=a2(1+cos⁡2θ−2cos⁡θ) ⇒x2=a2+a2cos⁡2θ−2a2cos⁡θ………(iii) \begin{aligned} & \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{aligned}
Taking the square of equation (ii)(ii), we get
y2=a2sin⁡2θ………(iv){{y}^{2}}={{a}^{2}}{{\sin }^{2}}\theta \ldots \ldots \ldots (iv)
Now, we can add equations (iii)(iii) and (iv)(iv).
x2+y2=a2+a2cos⁡2θ−2a2cos⁡θ+a2sin⁡2θ{{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
x2+y2=a2+a2cos⁡2θ+a2sin⁡2θ−2a2cos⁡θ x2+y2=a2+a2(cos⁡2θ+sin⁡2θ)−2a2cos⁡θ \begin{aligned} & {{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{aligned}
Since we know that sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1, we can apply the same in above equation and simplify as below,

& {{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{aligned}$$ Using the trigonometric relation, $1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2}$, we can simplify the above equation as, $$\begin{aligned} & {{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{aligned}$$ We can express the RHS as a square, $${{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