The position of a particle $P$ at time $t$ is given by $$x(t) = a \cos(\theta) i + b \sin(\theta) j$$
where $\theta = \theta(t)$ is a function, $a \neq 0$ and $b \neq 0$ are constants.
If the particle’s acceleration is always parallel to $x$ deduce that $\theta'$ is constant and that the direction of the acceleration is always from the particle towards the origin.
I have found $x''$ and the computed $x\times x''$ and equate it to zero hoping to find something but everything cancelled out.
I was thinking that if I can find $\theta''=0$ then that would integrate to $\theta'=\text{constant}$, but I do not know where to get $\theta''$ from.
$$\vec{x} = a\cos\theta \hat{i} + b\sin\theta\hat{j} $$
$$\frac{\vec{dx}}{dt} = \left(-a\sin\theta\hat{i}+ b\cos\theta \hat{j}\right)\frac{d\theta}{dt}$$
$$\frac{\vec{d^2x}}{dt^2}= \left(-a\sin\theta\hat{i}+ b\cos\theta \hat{j}\right)\frac{d^2\theta}{dt^2}-\left(a\cos\theta\hat{i}+ b\sin\theta \hat{j}\right)\frac{d\theta}{dt}$$
$$\vec{a} = \frac{\vec{d^2x}}{dt^2}= \left(-a\sin\theta\hat{i}+ b\cos\theta \hat{j}\right)\frac{d^2\theta}{dt^2}-\vec{x}\frac{d\theta}{dt}$$
We have $\vec{a} \parallel \vec{x} \implies \vec{a}\times\vec{x}=\vec{0}$
$\vec{a}\times \vec x= \left(a\cos\theta\hat{i}+ b\sin\theta \hat{j}\right)\times\left(-a\sin\theta\hat{i}+ b\cos\theta \hat{j}\right)\frac{d^2\theta}{dt^2}-\vec x \times\vec{x}\frac{d\theta}{dt} = \vec{0}$
$$(ab\cos^2\theta+ab\sin^2\theta)\frac{d^2\theta}{dt^2}\hat{k} = \vec{0}$$
$$ab\frac{d^2\theta}{dt^2}\hat{k} = \vec{0}$$
But $a,b\ne 0$
$$\implies \frac{d^2\theta}{dt^2} = 0$$