1. We are given two perpendicular SHMs:

$x(t)=A\sin(at+\delta), \qquad y(t)=B\sin(bt)$

We must identify which parameter set gives the stated Lissajous figure.

2. First, recall some standard results.

Case 1: Same frequency $a=b$

Then

$x=A\sin(\omega t+\delta), \qquad y=B\sin(\omega t)$

Eliminating $t$, the general curve is an ellipse:

$\frac{x^2}{A^2}+\frac{y^2}{B^2}-\frac{2xy}{AB}\cos\delta=\sin^2\delta$

This is the standard Lissajous figure for equal frequencies.

Special subcases:

• If $\delta=0$, then $x=A\sin\omega t, \quad y=B\sin\omega t$ so $x=\frac{A}{B}y$ which is a straight line.

• If $\delta=\frac{\pi}{2}$, then $x=A\cos\omega t, \quad y=B\sin\omega t$ so $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$ which is an ellipse. If additionally $A=B$, it becomes a circle.

3. Now check each option.

Option A

Parameters: $A\ne B$, $a=b$, $\delta=0$; Curve: Parabola

For $\delta=0$ and equal frequencies:

$x=A\sin\omega t, \qquad y=B\sin\omega t$

Hence

$x=\frac{A}{B}y$

This is a line, not a parabola.

So, A is incorrect.

Option B

Parameters: $A=B$, $a=b$, $\delta=\frac{\pi}{2}$; Curve: Line

For $\delta=\frac{\pi}{2}$ and equal frequencies:

$x=A\cos\omega t, \qquad y=A\sin\omega t$

Therefore,

$\frac{x^2}{A^2}+\frac{y^2}{A^2}=1$

or

$x^2+y^2=A^2$

This is a circle, not a line.

So, B is incorrect.

Option C

Parameters: $A\ne B$, $a=b$, $\delta=\frac{\pi}{2}$; Curve: Ellipse

For $\delta=\frac{\pi}{2}$ and equal frequencies:

$x=A\cos\omega t, \qquad y=B\sin\omega t$

Eliminating $t$:

$\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$

Since $A\ne B$, this is an ellipse.

So, C is correct.

Option D

Parameters: $A=B$, $a=2b$, $\delta=\frac{\pi}{2}$; Curve: Circle

Here frequencies are in ratio $2:1$, not equal. A circle requires equal frequencies with a phase difference of $\frac{\pi}{2}$ and equal amplitudes.

With $a=2b$,

$x=A\sin(2bt+\tfrac{\pi}{2})=A\cos 2bt, \qquad y=A\sin bt$

This gives a more complicated Lissajous figure, not a circle.

So, D is incorrect.

4. Therefore, the correct match is:

$\boxed{\text{C}}$

This matches the stored correct answer.