Mathematics - Application of Derivative Question with Solution | TestHub

Slope of line $(4-t)x+ty+\left(a^3-1\right)=0$ is $\frac{-(4-t)}{t}$ (or) $\frac{t-4}{t}$ $\because xy=1,\frac{dy}{dx}=\frac{-y}{x}=\frac{-1}{x^2};\therefore$ slope of normal $=x^2=\frac{t-4}{t}$ $\therefore x^2>0\Rightarrow \frac{t-4}{t}>0\Rightarrow t\in (-\infty ,0)\cup (4,\infty )$