If the line ` ax+by+c=0 ` is a tangent to the curve `xy=9,` then

Options

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

Explanation

To determine the condition under which the line \( ax + by + c = 0 \) is a tangent to the curve \( xy = 9 \), we can follow these steps:

### Step 1: Differentiate the curve
The given curve is \( xy = 9 \). We will differentiate this implicitly with respect to \( x \).

\[ \frac{d}{dx}(xy) = \frac{d}{dx}(9) \]
Using the product rule on the left side, we have:
\[ x \frac{dy}{dx} + y = 0 \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{y}{x} \]

### Step 2: Substitute for \( y \)
From the equation of the curve \( xy = 9 \), we can express \( y \) in terms of \( x \):
\[ y = \frac{9}{x} \]
Now, substituting this into the derivative:
\[ \frac{dy}{dx} = -\frac{\frac{9}{x}}{x} = -\frac{9}{x^2} \]

### Step 3: Find the slope of the line
The slope of the line given by the equation \( ax + by + c = 0 \) can be expressed as:
\[ \text{slope} = -\frac{a}{b} \]

### Step 4: Set the slopes equal
Since the line is a tangent to the curve, the slopes must be equal at the point of tangency:
\[ -\frac{9}{x^2} = -\frac{a}{b} \]
This simplifies to:
\[ \frac{9}{x^2} = \frac{a}{b} \]

### Step 5: Analyze the conditions
From the equation \( \frac{a}{b} = \frac{9}{x^2} \), we can deduce that for \( a \) and \( b \) to maintain the same sign, \( a \) and \( b \) must both be positive or both be negative. Thus, we conclude:
- If \( a > 0 \), then \( b > 0 \) (both must be positive).
- If \( a < 0 \), then \( b < 0 \) (both must be negative).

### Final Conclusion
The conditions for the line \( ax + by + c = 0 \) to be a tangent to the curve \( xy = 9 \) are:
- \( a > 0 \) and \( b > 0 \) or
- \( a < 0 \) and \( b < 0 \)