Question

Find the The Slopes of the tangent and the normal to the following curves at the indicated points : 

xy = 6 at (1, 6)

Answer 1

Given:

xy = 56 at (1,6)

Here we have to use the product rule for above equation.

If u and v are differentiable function, then

\(\therefore\frac{d}{dx}(Constant)=0\)

\(\Rightarrow x\frac{dy}{dx}+y=0\)

\(\Rightarrow x\frac{dy}{dx}=-y\)

\(\Rightarrow x\frac{dy}{dx}=\frac{-y}{x}\)

The Slope of the tangent at (1,6)is

\(\Rightarrow \frac{dy}{dx}=\frac{-6}{1}\)

\(\Rightarrow \frac{dy}{dx}=-6\)

\(\therefore\) The Slope of the tangent at (1,6) is – 6

⇒ The Slope of the normal =\(\frac{-1}{\text{The Slope of the tangent}}\)

⇒ The Slope of the normal = \(\frac{-1}{(\frac{dy}{dx})}\)

⇒ The Slope of the normal =\(\frac{-1}{-6}\)

⇒ The Slope of the normal = \(\frac{1}{6}\)