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If the tangent to the curve $x y+a x+b y=0$ at $(1,1)$ is inclined at an angle $\tan ^{-1} 2$ with $X$-axis, then
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Verified Answer
The correct answer is:
$a=1, b=-2$
Given curve, $x y+a x+b y=0$
$(1,1)$ lie on curve (i)
$$
\begin{aligned}
& \therefore \quad 1+a+b=0 \\
& \text { From Eq. (i), } \\
& x \frac{d y}{d x}+y \cdot 1+a+b \frac{d y}{d x}=0 \\
& \Rightarrow \quad \frac{d y}{d x}=\frac{-(a+y)}{(b+x)} \\
& \therefore\left(\frac{d y}{d x}\right)_{(1,1)}=-\frac{(a+1)}{b+1}=2 \quad(\because m=\tan \theta) \\
& \therefore \quad \frac{-(-b)}{b+1}=2 \\
& \Rightarrow \quad b=2 b+2 \\
& \therefore \quad b=-2 \\
& \text { From Eq. (ii), } a=1 \\
& \text { Hence, } a=1, b=-2 \text {. } \\
&
\end{aligned}
$$
[from Eq. (ii)]
$(1,1)$ lie on curve (i)
$$
\begin{aligned}
& \therefore \quad 1+a+b=0 \\
& \text { From Eq. (i), } \\
& x \frac{d y}{d x}+y \cdot 1+a+b \frac{d y}{d x}=0 \\
& \Rightarrow \quad \frac{d y}{d x}=\frac{-(a+y)}{(b+x)} \\
& \therefore\left(\frac{d y}{d x}\right)_{(1,1)}=-\frac{(a+1)}{b+1}=2 \quad(\because m=\tan \theta) \\
& \therefore \quad \frac{-(-b)}{b+1}=2 \\
& \Rightarrow \quad b=2 b+2 \\
& \therefore \quad b=-2 \\
& \text { From Eq. (ii), } a=1 \\
& \text { Hence, } a=1, b=-2 \text {. } \\
&
\end{aligned}
$$
[from Eq. (ii)]
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