The X and y-coordinates of a particle moving in a plane are given by x(t) = a cos(pt) and y(t) = b sin(pt), where a , b ($<$ a) and p are positive constants of appropriate dimensions and t is time. Then, which of the following is not true?

Given, x = a cos(pt), y = b sin(pt)

$\therefore \frac{x^{\mathit{2}}}{a^{\mathit{2}}} + \frac{y^{\mathit{2}}}{b^{\mathit{2}}} = 1$, i.e. equation of ellipse

Now, $\mathbf{r}=\hat{\mathbf{i}}+\hat{\mathbf{j}}=a\cos \left(pt\right)\hat{\mathbf{i}}+b\sin \left(pt\right)\hat{\mathbf{j}}$

         $\mathbf{v}=\frac{d\mathbf{r}}{dt}=-pa\sin \left(pt\right)\hat{\mathbf{i}}+pb\cos \left(pt\right)\hat{\mathbf{j}}$

         $\mathbf{a}=\frac{dv}{dt}=-p^{2}a\cos \left(pt\right)\hat{\mathbf{i}}-p^{2}b\sin \left(pt\right)\hat{\mathbf{j}}$

At      t = $\frac{\mathrm{\pi }}{2p}$,

        $\mathbf{v}=-pa\hat{\mathbf{i}}\text{ and }\mathbf{a}=-p^2\stackrel{ }{b}\hat{ȷ}$

Thus, velocity is perpendicular to acceleration. Also, $a = -p^{2}r$, i.e. directed towards a fixed point as it is always opposite to position vector.

$\text{ As, } v=\frac{dr}{dt}\Rightarrow \Delta r={\int }_0^{\frac{\pi }{2p}}vdt$

$\Rightarrow \Delta r=\left[a\cos \right(pt)\hat{\mathbf{i}}+b\sin (pt)\hat{\mathbf{j}}{]}_0^{\frac{\pi }{2p}}=-a\hat{\mathbf{i}}+b\hat{\mathbf{j}}$

$\text{ So, } \left|\mathbf{r}\right|=\sqrt{a^2+b^2}$