If slope of the tangent to the curve $x y+a x+b y=0$ at the point $(1,1)$ on it is 2 , then the value of $3 a+b$ is (

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Aicharya · Accepted Answer

$x y+a x+b y=0$ Differentiating w.r.t. $x$, we get $\frac{x \mathrm{~d} y}{\mathrm{~d} x}+y+\mathrm{a}+\mathrm{b} \frac{\mathrm{d} y}{\mathrm{~d} x}=0$ $\begin{aligned} & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{\mathrm{a}+y}{\mathrm{~b}+x} \ & \Rightarrow\left(\frac{\mathrm{d} y}{\mathrm{~d} x}ight)_{(\mathrm{b}, 1)}=-\frac{\mathrm{a}+1}{\mathrm{~b}+1} \ & \Rightarrow 2=-\frac{\mathrm{a}+1}{\mathrm{~b}+1}\end{aligned}$ $\Rightarrow a+2 b=-3$ ...(i) Also, the point $(1,1)$ lies on the curve $x y+a x+b y=0$. $	herefore \quad 1+a+b=0$ $\Rightarrow a+b=-1$ ...(ii) Solving (i) and (ii), we get $a=1, b=-2$ $	herefore \quad 3 a+b=1$

If slope of the tangent to the curve $x y+a x+b y=0$ at the point $(1,1)$ on it is 2 , then the value of $3…

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