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Idea of parametric form of differentiation:

If $y=f(\theta)$ and $x=g(\theta)$ i.e. $y$ is a function of $\theta$ and $x$ is also some other function of $\theta$.

Then $d y / d \theta=f^{\prime}(\theta)$ and $d x / d \theta=g^{\prime}(\theta)$

We can write : $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$

Given,

$x=a(\theta+\sin \theta) \ldots \ldots$ equation 1

$y=a(1+\cos \theta) \ldots \ldots$ equation 2

to prove : $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{\mathrm{a}}{\mathrm{y}^{2}}$

We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.

Let's find $\frac{d^{2} y}{d x^{2}}$

$\mathrm{AS}, \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

So, lets first find $d y / d x$ using parametric form and differentiate it again.

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\frac{\mathrm{d}}{\mathrm{d} \theta} \mathrm{a}(\theta+\sin \theta)=\mathrm{a}(1+\cos \theta)=\mathrm{y}[\because$ from equation 2$] \ldots . .$ equation 3

Similarly,

$\frac{\mathrm{dy}}{\mathrm{d} \theta}=\frac{\mathrm{d}}{\mathrm{d} \theta} \mathrm{a}(1+\cos \theta)=-\mathrm{asin} \theta \ldots \ldots .$ equation 4

$\left[\because \frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}=-\sin \mathrm{x}, \frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}=\cos \mathrm{x}\right.$

$\left[\because \frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}=-\sin \mathrm{x}, \frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}=\cos \mathrm{x}\right.$

Differentiating again w.r.t $\mathrm{x}$ :

$\frac{d}{d x}\left(\frac{d y}{d x}\right)=-a \frac{d}{d x}\left(\frac{\sin \theta}{y}\right)$

Using product rule and chain rule of differentiation together:

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\mathrm{a}\left(\frac{\sin \theta}{-\mathrm{y}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{1}{\mathrm{y}} \cos \theta \frac{\mathrm{d} \theta}{\mathrm{dx}}\right)$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\mathrm{a}\left(\frac{\sin \theta}{-\mathrm{y}^{2}} \frac{(-a \sin \theta)}{\mathrm{y}}+\frac{1}{\mathrm{y}} \cos \theta \frac{1}{\mathrm{y}}\right)$ [using equation 3 and 5]

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\mathrm{a}\left(\frac{\mathrm{asin}^{2} \theta}{\mathrm{y}^{3}}+\frac{1}{\mathrm{y}^{2}} \cos \theta\right)$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{\mathrm{a}}{\mathrm{y}^{2}}\left(\frac{\mathrm{a} \sin ^{2} \theta}{\mathrm{a}(1+\cos \theta)}+\cos \theta\right)[$ from equation 1]

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{\mathrm{a}}{\mathrm{y}^{2}}\left(\frac{1-\cos ^{2} \theta}{(1+\cos \theta)}+\cos \theta\right)$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{\mathrm{a}}{\mathrm{y}^{2}}\left(\frac{(1-\cos \theta)(1+\cos \theta)}{(1+\cos \theta)}+\cos \theta\right)$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{\mathrm{a}}{\mathrm{y}^{2}}(1-\cos \theta+\cos \theta)$

$\therefore \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{\mathrm{a}}{\mathrm{y}^{2}} \ldots$ proved