Given $\alpha$ and $\beta$ are roots of $x^2+ax-b=0$

$\alpha +\beta =-a ...\left(1\right)$

$\alpha \beta =-b ...\left(2\right)$

Now, differentiating the curve ${\left(\frac{x}{\alpha }\right)}^n+{\left(\frac{y}{\beta }\right)}^n=2$

$\frac{n}{\alpha }{\left(\frac{x}{\alpha }\right)}^{n-1}+\frac{n}{\beta }{\left(\frac{y}{\beta }\right)}^{n-1}\frac{dy}{dx}=0$

at $x=\alpha$ and $y=\beta$

$\frac{dy}{dx}=\frac{-\beta }{\alpha } ...\left(3\right)$

Now, $x\cos \theta +y\sin \theta =c$

$y=-xcot\theta +\frac{c}{\sin \theta }$

As, $-cot\theta =\frac{-\beta }{\alpha }$

$y=\frac{-\beta }{\alpha }x+\frac{c}{\sin \theta }$

Put $=\left(\alpha , \beta \right)$

$\Rightarrow \beta =\frac{c}{2\sin \theta }$ and $\alpha =\frac{c}{2\cos \theta } ...\left(4\right)$

Now, ${\left(\frac{a}{b}\right)}^2+\frac{2}{b}=$

By Equations $\left(1\right) \& \left(2\right)$

$\Rightarrow {\left(\frac{-\left(\alpha +\beta \right)}{-\left(\alpha \beta \right)}\right)}^2-\frac{2}{\alpha \beta }$

$\Rightarrow \frac{{\alpha }^2+{\beta }^2+2\alpha \beta -2\alpha \beta }{(\alpha \beta {)}^2}$

$\Rightarrow \frac{{\alpha }^2+{\beta }^2}{{\alpha }^2{\beta }^2}$

By Equation $\left(4\right)$

$\Rightarrow \frac{\frac{c^2}{4{\cos }^2\theta }+\frac{c^2}{4{\sin }^2\theta }}{{\displaystyle \frac{c^2\times c^2}{4{\cos }^2\theta 4{\sin }^2\theta }}}$

$\Rightarrow \frac{{\displaystyle \frac{c^2}{4}}\left({\displaystyle \frac{{\cos }^2\theta +{\sin }^2\theta }{{\cos }^2\theta {\sin }^2\theta }}\right)}{{\displaystyle \frac{c^4}{16{\cos }^2\theta {\sin }^2\theta }}}\Rightarrow \frac{4}{c^2}$

${\left(\frac{a}{b}\right)}^2+\frac{2}{b}=\frac{4}{c^2}$