The coordinates of a particle moving in a plane are given by $x = a \cos (p t)$ and $y = b \sin (p t)$ , where $a, b (< a)$ and $p$ are positive constants of appropriate dimensions, then

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The acceleration of the particle is always directed towards a focus

The distance travelled by the particle in time interval, $t = 0$ to $t = π / 2 p$ is $a$

The velocity and acceleration of the particle are normal to each other at $t = π / 2 p$

The path of the particle is an ellipse

answer is A, B.

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Detailed Solution

$x = a \cos (p t), y = b \sin (p t)$
Equation of path in x-y plane: ${[\frac{x}{a}]}^{2} + {[\frac{y}{a}]}^{2} = 1$ i.e., the path of the particle is an ellipse
Position vector of a point P is: $\vec{r} = a \cos p t \hat{i} + b \sin p t \hat{j}$
$\Rightarrow \vec{v} = p (- a \sin p t \hat{i} + b \cos p t \hat{j})$ and $\vec{a} = - p^{2} (a \cos p t \hat{i} + b \sin p t \hat{j}) = - p^{2} \vec{r}$
Also $\vec{v} . \vec{a} = 0$ at $t = π / 2 p$