If the line ax + by + c =0, ab `ne` 0, is a tangent to the curve xy =1-2x, then

Options

A. `a gt 0, b lt 0`
B. `a gt 0, b gt0`
C. `a lt 0, b gt 0`
D. `a lt 0, b lt 0`

Answer

To solve the problem, we need to determine the conditions on the coefficients \( a \) and \( b \) of the line \( ax + by + c = 0 \) given that this line is a tangent to the curve defined by \( xy = 1 - 2x \).

### Step 1: Differentiate the curve equation
We start with the equation of the curve:
\[ xy = 1 - 2x \]
To find the slope of the tangent line at any point on the curve, we differentiate both sides with respect to \( x \):
\[ y + x \frac{dy}{dx} = -2 \]
Rearranging gives:
\[ x \frac{dy}{dx} = -2 - y \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{-2 - y}{x} \]

### Step 2: Find the slope of the tangent line
The slope of the line \( ax + by + c = 0 \) can be expressed in terms of \( a \) and \( b \):
\[ \text{slope} = -\frac{a}{b} \]
Since the line is tangent to the curve, the slopes must be equal at the point of tangency:
\[ -\frac{a}{b} = \frac{-2 - y}{x} \]

### Step 3: Substitute \( y \) from the curve equation
From the curve equation \( xy = 1 - 2x \), we can express \( y \) in terms of \( x \):
\[ y = \frac{1 - 2x}{x} \]
Now substituting this expression for \( y \) into the slope equation:
\[ -\frac{a}{b} = \frac{-2 - \frac{1 - 2x}{x}}{x} \]
Simplifying the right-hand side:
\[ -\frac{a}{b} = \frac{-2 - \frac{1}{x} + 2}{x} = \frac{-\frac{1}{x}}{x} = -\frac{1}{x^2} \]
Thus we have:
\[ \frac{a}{b} = \frac{1}{x^2} \]

### Step 4: Analyze the sign of \( a \) and \( b \)
Since \( x^2 > 0 \) for all \( x \neq 0 \), it follows that \( \frac{a}{b} > 0 \). This implies that both \( a \) and \( b \) must either be both positive or both negative.

### Conclusion
Thus, we conclude that:
- If \( a > 0 \), then \( b > 0 \).
- If \( a < 0 \), then \( b < 0 \).

### Final Answer
The relationship between \( a \) and \( b \) is such that \( \frac{a}{b} > 0 \).