Normal line to a curve $C_1$ > Find the interval for $a$ so that $(3-a)x+ay+(a^2-1)=0$ is normal to the curve $xy=4$ $(C_1)$. I approached it this way-- $C_1$ is $xy=4$. So, $\dfrac{dy}{dx}$ for $C_1$ is $\dfrac{-4}{x^2}$. So, the slope of the normal drawn at a point $\left(x_1,\dfrac{4}{x_1}\right)$ to the curve $C_1$ would be $\dfrac{(x_1)^2}{4}$. Let the line be normal to the curve at $\left(x_1,\dfrac{4}{x_1}\right)$. So, two conditions should be satisfied. $1)$ The slope of the normal to the curve should be equal to the slope of the line. $2)$ The line should pass pass through that point. Writing these two quantitatively, $$\rightarrow\dfrac{a-3}{a}=\dfrac{x_1^2}{4}$$ $$\rightarrow (3-a)x_1+\dfrac{4a}{x_1}+(a^2-1)=0$$ Solving these two for a yields $\boxed{a=4}$ and $\boxed{a \approx -3.679}$. But the answer says $a \in (-\infty,0) \cup (3,\infty)$. Where did i go wrong?

Your approach is correct, but the problem does not require that the two functions are normal in the crossing point. We have only to determine for which values of $a$ there exists a point where the slope of the curve $y=4/x$ is normal to that of the line $y=\frac{a-3}{a}x-\frac{a^2-1}{a}$. Thus we can focus on the first of the two equations correctly shown in the question, i.e.

$$\frac{a-3}{a}=\frac{x^2}{4}$$

from which

$$x=\sqrt{\frac{4(a-3)}{a}}$$

The last equation has real solutions only for $\frac{a-3}{a}\geq 0$, so that we get $a< 0$ or $a>3$. Note that the values $a=3$ and $a=0$ have to be excluded because the slope of the line reduces to $0$ and $\infty$, respectively.