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 find `aa n db ?`

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript:

### Step 1: Differentiate the curve
The given curve is:
\[ xy + ax + by = 0 \]

We differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(xy) + \frac{d}{dx}(ax) + \frac{d}{dx}(by) = 0 \]
Using the product rule on \( xy \):
\[ x \frac{dy}{dx} + y + a + b \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ (x + b) \frac{dy}{dx} + y + a = 0 \]
Thus,
\[ \frac{dy}{dx} = -\frac{y + a}{x + b} \]

### Step 2: Find the slope at the point (1, 1)
We need to find the slope of the tangent at the point \( (1, 1) \):
\[ \frac{dy}{dx}\bigg|_{(1,1)} = -\frac{1 + a}{1 + b} \]

### Step 3: Set the slope equal to the tangent of the angle
The angle of inclination is given as \( \tan^{-1}(2) \), which means:
\[ \tan(\theta) = 2 \]
Thus, we have:
\[ -\frac{1 + a}{1 + b} = 2 \]
This leads to the equation:
\[ 1 + a = -2(1 + b) \]
Expanding this gives:
\[ 1 + a = -2 - 2b \]
Rearranging gives:
\[ a + 2b = -3 \quad \text{(Equation 1)} \]

### Step 4: Substitute point (1, 1) into the original curve
Now we substitute \( x = 1 \) and \( y = 1 \) into the original curve:
\[ 1 \cdot 1 + a \cdot 1 + b \cdot 1 = 0 \]
This simplifies to:
\[ 1 + a + b = 0 \]
Rearranging gives:
\[ a + b = -1 \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations
Now we have two equations:
1. \( a + 2b = -3 \)
2. \( a + b = -1 \)

We can subtract Equation 2 from Equation 1:
\[ (a + 2b) - (a + b) = -3 - (-1) \]
This simplifies to:
\[ b = -2 \]

### Step 6: Substitute back to find \( a \)
Now substituting \( b = -2 \) into Equation 2:
\[ a + (-2) = -1 \]
This gives:
\[ a - 2 = -1 \implies a = 1 \]

### Final Result
Thus, the values of \( a \) and \( b \) are:
\[ a = 1, \quad b = -2 \]