Two simple harmonic motions, as shown, are at right angles. They are combined to form Lissajous figures.
\[x\left( t \right) = Asin\left( {at + \delta } \right)\]
\[y\left( t \right) = Bsin\left( {bt} \right)\]
Identify the correct match below
A. Parameters: \[A = B,a = 2b;\delta = \dfrac{\pi }{2}\]; Curve: Circle
B. Parameters: \[A = B,a = b;\delta = \dfrac{\pi }{2}\]; Curve: Line
C. Parameters \[A \ne B,a = b;\delta = \dfrac{\pi }{2}\]; Curve: Ellipse
D. Parameters:\[A \ne B,a = b;\delta = 0\]; Curve: Parabola

Hint: This question is solved using the trial and error method. By assuming the different data given in the options, we will solve the two wave equations until a particular equation of a closed geometrical shape arises.

Formula Used: Harmonic motion wave equation: \[x\left( t \right) = Asin\left( {at + \delta } \right)\], \[y\left( t \right) = Bsin\left( {bt} \right)\]

Where $x,y$ are the wave paths of the two harmonic waves, $t$ is the time taken for the wave to oscillate, $A,B$ are the amplitudes of the two harmonic waves, $a,b$ are the phases of the two harmonic waves and $\delta $ is the phase difference between the two waves.

Complete step by step answer:
This question will be solved using trial and error method. That means, we will assume the data given in each option as true to the wave equations and substitute them in it. This will help us find out which set of data is true.
We will be taking the data given in option C to be true.
As given in the option we have the following information:
Amplitude: \[A \ne B\]
Phase: \[a = b\]
Phase difference: \[\delta = \dfrac{\pi }{2}\]
We have \[x\left( t \right) = Asin\left( {at + \delta } \right)\]. When we expand the $\sin $ part we get,
\[x\left( t \right) = Asin\left( {at + \delta } \right) = A[\sin (at).\cos \delta + \sin \delta .\cos (at)]\]
Simplifying this we get,
\[
  x\left( t \right) = A[\sin (at).\cos \delta + \sin \delta .\cos (at)] \\
   \Rightarrow \dfrac{{x(t)}}{A} = [\sin (at).\cos \delta + \sin \delta .\cos (at)] \\
 \]
We determine the value of \[\dfrac{{x(t)}}{A}\] from the second wave equation, which is \[y\left( t \right) = Bsin\left( {bt} \right)\]. But we know that $a = b$.
Therefore we get,
\[
  y\left( t \right) = Bsin\left( {bt} \right) = Bsin\left( {at} \right) \\
   \Rightarrow sin\left( {at} \right) = \dfrac{{y\left( t \right)}}{B} \\
 \]
According to sine-cosine trigonometrical theorem, if \[sin\left( {at} \right) = \dfrac{{y\left( t \right)}}{B}\], then \[\cos (at)\] will be equal to \[\sqrt {1 - {{(\dfrac{{y\left( t \right)}}{B})}^2}} \].
Substituting this value of $\sin (at)$ in our expanded first wave equation we get,
\[
  \dfrac{{x(t)}}{A} = [\sin (at).\cos \delta + \sin \delta .\cos (at)] = [\dfrac{{y\left( t \right)}}{B}.\cos \delta + \sin \delta .\sqrt {1 - {{(\dfrac{{y\left( t \right)}}{B})}^2}} ] \\
  \dfrac{{x(t)}}{A} = [\dfrac{{y\left( t \right)}}{B}.\cos \delta + \sin \delta .\sqrt {1 - {{(\dfrac{{y\left( t \right)}}{B})}^2}} ] \\
 \]

Simplifying this we will get,
$\dfrac{{{x^2}}}{{{A^2}}} + \dfrac{{{y^2}}}{{{B^2}}} - \dfrac{{2xy}}{{AB}}\cos \delta = {\sin ^2}\delta $
But we know that \[\delta = \dfrac{\pi }{2}\]. At this angle $\sin \delta $ will be max, that is $1$, if \[\delta = \dfrac{\pi }{2}\] and \[\cos \delta = 0\].
Therefore the final equation turns out to be $\dfrac{{{x^2}}}{{{A^2}}} + \dfrac{{{y^2}}}{{{B^2}}} = 1$.
This is the equation of an eclipse. Also it was determined using \[a = b\] and \[\delta = \dfrac{\pi }{2}\]. Therefore satisfies them both. We also observe that \[A \ne B\].

In conclusion, the correct option is A.

Note:If this question is not solved using trial and error method, it will be difficult to determine the answer. We need to start with a certain set of data. By solving for each we will understand which set of given data is correct.