#### Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 1 Sub Question 10 Maths Textbook Solution.

$\text { The slope of the tangent is }-6$

$\text { The slope of the normal is } \frac{1}{6}$

Hint:

$\text { The slope of tangent is the limit of } \frac{\Delta y}{\Delta x} \text { as } \Delta x \text { approaches zero. }$

$\text { The slope of normal is the normal to a curve at } P=(x, y) \text { is a line perpendicular to the tangent }$

$\text { at } P \text { and passing through } P \text { . }$

Given:$x y=6 \text { at }(1,6)$

$\text { Here we have to use product rule, }$

Solution:

$\frac{d}{d x}(x y)=\frac{d}{d x}(6)$

$x \frac{d}{d x}(y)+y \frac{d}{d x}(x)=\frac{d}{d x}(6)$

$\frac{d}{d x} \text { (constant) }=0$

$x \frac{d y}{d x}+y=0$

$x \frac{d y}{d x}=-y$

$\Rightarrow \frac{d y}{d x}=\frac{-y}{x}$

$\text { The slope of tangent at }(1,6) \text { is }$

$\frac{d y}{d x}=\frac{-6}{1}=-6$

$\text { The slope of the tangent at }(1,6) \text { is }-6$

$\text { The slope of the normal }=\frac{-1}{\text { The slope of the tangent }}$

$\text { The slope of the normal }=\frac{-1}{\left(\frac{d y}{d x}\right)}$

$\text { The slope of the normal }=\frac{-1}{-6}=\frac{1}{6}$