Step 1: Understanding the Question:
A line is a normal to a hyperbola. We need to find the relationship between the coefficients \( a \) and \( b \).
Step 2: Key Formula or Approach:
The slope of the normal is the negative reciprocal of the derivative of the curve.
For a curve \( y = f(x) \), slope of tangent \( m_t = \frac{dy}{dx} \) and slope of normal \( m_n = -\frac{1}{dy/dx} \).
Step 3: Detailed Explanation:
The curve is \( xy = 1 \implies y = \frac{1}{x} \).
Differentiating with respect to \( x \):
\[ \frac{dy}{dx} = -\frac{1}{x^2} \]
The slope of the normal at any point \( (x, y) \) is:
\[ m_n = -\frac{1}{-1/x^2} = x^2 \]
Since \( x^2 \) is always strictly positive for any real \( x \ne 0 \), the slope of the normal to this curve must be positive.
The equation of the given line is \( ax + by + c = 0 \), which can be written as \( y = -\frac{a}{b}x - \frac{c}{b} \).
The slope of this line is \( m = -\frac{a}{b} \).
For this line to be a normal, we must have:
\[ -\frac{a}{b}>0 \]
This inequality holds if \( a \) and \( b \) have opposite signs.
From the options:
(A) \( a, b \) both positive \( \implies \) slope is negative.
(B) \( a>0, b<0 \implies -\frac{(+)}{(-)} = (+) \). This is a positive slope.
(C) \( a<0, b \ge 0 \). If \( b=0 \), the line is vertical (not defined by slope).
(D) \( a, b \) both negative \( \implies \) slope is negative.
Thus, option (B) is the most suitable general choice.
Step 4: Final Answer:
The correct condition is \( a>0, b<0 \).