If the tangent to the curve xy+ ax+ by= 0 at (1, 1) is inclined at an angle tan−1⁡2 with x-axis, then

A
$a = 1, b = 2$
B
$a = 1, b = - 2$
C
$a = - 1, b = 2$
D
$a = - 1, b = - 2$

Solution:

The point $(1, 1)$ lies on the curve $x y + a x + b y = 0$

$∴ a + b = - 1$

Now, $x y + a x + b y = 0$

$\begin{aligned} \Rightarrow x \frac{d y}{d x} + y + a + b \frac{d y}{d x} = 0 \\ \Rightarrow {(\frac{d y}{d x})}_{(1, 1)} = - \frac{a + 1}{b + 1} \end{aligned}$

Since the tangent makes an angle $\tan^{- 1} 2$ with x-axis.

Slope of the tangent = 2

$\Rightarrow 2 = - \frac{a + 1}{b + 1} \Rightarrow a + 2 b = - 3$

Solving (i) and (iii), we get $a = 1, b = - 2$

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

Solution:

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