Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 1 Sub Question 9 Maths Textbook Solution.

$\text { The slope of the tangent is }-\frac{2}{5}$

$\text { The slope of the normal is } \frac{5}{2}$

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:$x^{2}+3 y+y^{2}=5 \text { at }(1,1)$

$\text { Here we have to differentiate the equation with respect to } x$

Solution:

$\frac{d}{d x}\left(x^{2}+3 y+y^{2}\right)=\frac{d}{d x}(5)$

$\frac{d}{d x}\left(x^{2}\right)+\frac{d}{d x}(3 y)+\frac{d}{d x}\left(y^{2}\right)=\frac{d}{d x}(5)$

$\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$

$\Rightarrow 2 x+3 \frac{d y}{d x}+2 y \frac{d y}{d x}=0$

$\Rightarrow 2 x+\frac{d y}{d x}(3+2 y)=0$

$\Rightarrow \frac{d y}{d x}(3+2 y)=-2 x$

$\frac{d y}{d x}=\frac{-2 x}{(3+2 y)}$

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

$\frac{d y}{d x}=\frac{-2 \times 1}{(3+2 \times 1)}=\frac{-2}{3+2}$

$\frac{d y}{d x}=\frac{-2}{5}$

$\text { The slope of the tangent at }(1,1) \text { is } \frac{-2}{5}$

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

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

$\text { The slope of the normal }=\frac{-1}{\frac{-2}{5}}=\frac{5}{2}$