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#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 2 Maths Textbook Solution.

Answer:$a=5 \text { and } b=-4$

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

Given:

$\text { The slope of the tangent to the curve } x y+a x+b y=2 \text { at }(1,1) \text { is } 2$

Solution:

$\text { First we will find the slope of tangent by using product rule, we get }$

$x y+a x+b y=2$

$x \frac{d}{d x}(y)+y \frac{d}{d x}(x)+a \frac{d}{d x}(x)+b \frac{d}{d x}(y)=\frac{d}{d x}(2)$

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

$\frac{d y}{d x}(x+b)=-(a+y)$

$\frac{d y}{d x}=\frac{-(a+y)}{(x+b)}$

$\text { The slope of tangent to the curve } x y+a x+b y=2 \text { at }(1,1) \text { is } 2$

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

$\Rightarrow \frac{-(a+y)}{(x+b)}=2$

$\Rightarrow \frac{-(a+1)}{(1+b)}=2$

$\Rightarrow-(a+1)=2(1+b)$

$\Rightarrow-(a+1)=2+2 b$

$\Rightarrow a+2 b=-3 \quad \rightarrow(1)$

$\text { Also the point }(1,1) \text { lies on the curve } x y+a x+b y=2 \text { we have, }$

\begin{aligned} &1 \times 1+a \times 1+b \times 1=2 \\ &1+a+b=2 \\ &a+b=1 \quad \rightarrow(2) \end{aligned}

\begin{aligned} &\text { Subtract eqn(1) and (2) }\\ &a+2 b=-3\\ &\frac{a+b=1}{b=-4} \end{aligned}

$\text { Substitute } b=-4 \text { in } a+b=1$

\begin{aligned} &a-4=1 \\ &a=5 \\ &\text { So, } a=5 \text { and } b=-4 \end{aligned}

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