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

If slope of a tangent to the curve $x y+a x+b y=0$ at the point $(1,1)$ on it is 2 , then $a-b$ is
  1. 3
  2. 1
  3. 2
  4. -1

Solution

Given curve is $\begin{aligned} & x y+\mathrm{a} x+\mathrm{b} y=0 \\ \therefore \quad & \text { Slope }=2=\frac{\mathrm{d} y}{\mathrm{~d} x} \\ \therefore \quad & x y+\mathrm{a} x+\mathrm{b} y=0 \end{aligned}$ Differentiating w.r.t. $x$, we get $\begin{aligned} & x \frac{\mathrm{d} y}{\mathrm{~d} x}+y+\mathrm{a}+\mathrm{b} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\ & \therefore \quad(x+\mathrm{b}) \frac{\mathrm{d} y}{\mathrm{~d} x}=-(y+\mathrm{a}) \\ & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-(y+\mathrm{a})}{x+\mathrm{b}} \\ & \therefore \quad\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(1,1)}=2 \\ & \therefore \quad 2=\frac{-(1+\mathrm{a})}{1+\mathrm{b}} \\ & \therefore \quad \mathrm{a}+2 \mathrm{~b}=-3 ... (i) \end{aligned}$ Since $(1,1)$ lies on $x y+a x+b y=0$, we get $\mathbf{a}+\mathbf{b}=-1$ ... (ii) Solving (i), (ii), we get $\begin{aligned} & a=1, b=-2 \\ \therefore \quad & a-b=1-(-2)=3 \end{aligned}$

Asked in: MHT CET 2023 (09 May Shift 2)

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