Let $y=f\left(x\right)$ be a curve $C_1$ passing through $(2,2)$ and $(8,\frac{1}{2})$ and satisfying a differential equation $y\left(\frac{d^{2}y}{d{x}^2}\right)=2{\left(\frac{dy}{dx}\right)}^2.$ Curve $C_2$ is the director circle of the circle $x^2+y^2=2.$ If the shortest distance between the curves $C_1$ and $C_2$ is $(\sqrt{p}-q)$ where $p,q\in N,$ then find the value of $(p^2-q).$

Given $y\left(\frac{d^{2}y}{d{x}^2}\right)=2{\left(\frac{dy}{dx}\right)}^2\Rightarrow \frac{y^{\text{''}}}{y^{\text{'}}}=\frac{2y^{\text{'}}}{y}\Rightarrow \text{ln}{y}^{\text{'}}=2\text{ln}y+\text{ln}a$

$\Rightarrow \frac{y^2}{y^2}=a\Rightarrow {\int }^{\text{}}\frac{dy}{y^2}={\int }^{\text{}}adx\Rightarrow \frac{-1}{y}=ax+b$

Since curve is passing through (2,2) and $(8,\frac{1}{2})$

So, $2a+b=\frac{-1}{2}$

$8a+b=-2$

$\therefore$On solving (i) and (ii), we get $a=\frac{-1}{4},b=0$

Hence, $C_1:xy=4$ and curve $C_2$ is $x^2+y^2=4.$

$\therefore$Shortest distance between $C_1$ and $C_2=2\sqrt{2}-2=\sqrt{8}-2$ 

Hence, $(p^2-q)=64-2=62$