Question

Find the slope of the tangent and the normal to the following curve at the indicted point  xy = 6 at (1, 6) ?

Answer 1

$$xy=6$$

$$\text{ On differentiating both sides w.r.t. x, we get }$$

$$x\frac{dy}{dx}+y=0$$

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

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

$$\text{ Now,}$$

$$\text{ Slope of the tangent }={\left(\frac{dy}{dx}\right)}_{(1,6)}=\frac{-y}{x}=\frac{-6}{1}=-6$$

$$\text{ Slope of the normal }=\frac{-1}{{\left(\frac{dy}{dx}\right)}_{(1,6)}}=\frac{-1}{-6}=\frac{1}{6}$$