The value of the parameter ‘$a$’ so that the line $\left(3-a\right)x+ay+\left(a^2-1\right)=0$ is normal to the curve $xy=1$may lie in the interval

 the line $\left(3-a\right)x+ay+\left(a^2-1\right)=0$ is normal to the curve $xy=1$

$y=\frac{1}{x}\Rightarrow \frac{dy}{dx}=-\frac{1}{x^2}$

Slope of normal =$\frac{-1}{{\displaystyle \frac{dy}{dx}}}=x_1^2>0$

$\text{ Slope of given line is }\frac{a-3}{a}\ge 0$

$\therefore \frac{a-3}{a}>0$

$\Rightarrow a\in (-\mathrm{\infty },0)\cup (3,\mathrm{\infty })$