# Find the values of a and b

Question:

Find the values of $a$ and $b$ if the The Slope of the tangent to the curve $x y+a x+b y=2$ at $(1,1)$ is 2 .

Solution:

Given:

The Slope of the tangent to the curve $x y+a x+b y=2$ at $(1,1)$ is 2

First, we will find The Slope of tangent

we use product rule here,

$\therefore \frac{\mathrm{d}}{\mathrm{dx}}(U V)=U \times \frac{\mathrm{dV}}{\mathrm{dx}}+\mathrm{V} \times \frac{\mathrm{dU}}{\mathrm{dx}}$

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

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

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

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

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

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

since, The Slope of the tangent to the curve $x y+a x+b y=2$ at $(1,1)$ is 2

i.e, $\frac{\mathrm{dy}}{\mathrm{dx}}=2$

$\Rightarrow\left\{\frac{-(a+y)}{x+b}\right\}(x=1, y=1)=2$

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

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

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

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

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

$1 \times 1+a \times 1+b \times 1=2$

$\Rightarrow 1+a+b=2$

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

from $(1) \&(2)$, we get

\begin{tabular}{c}

$a+2 b=-3$ \\

$a+b=1$ \\

$-\quad-\quad-$ \\

\hline$b=-4$

\end{tabular}

substitute $b=-4$ in $a+b=1$

$a-4=1$

$\Rightarrow a=5$

So the value of $a=5 \& b=-4$