Question

If the line \( ax + by + c = 0 \) is normal to the curve \( xy = 1 \), then}

• A. \( a>0, b>0 \)
• B. \( a>0, b<0 \)
• C. \( a<0, b \ge 0 \)
• D. \( a<0, b<0 \)

Hint:

The slope of the normal to $xy=c$ is always positive because the tangent slope $-c/x^2$ is always negative.

Answer 1

The Correct Option is B

Step 1: Slope of Tangent
$xy = 1 \implies y = 1/x$.
$dy/dx = -1/x^2$. 
Step 2: Slope of Normal
$m_{normal} = -1 / (dy/dx) = x^2$. 
Since $x$ is real, $x^2>0$. The slope of the normal must be positive. 
Step 3: Analyze Line Slope
$ax + by + c = 0 \implies y = (-a/b)x - c/b$. 
Slope $m = -a/b$. 
For $m>0$, $-a/b>0 \implies a/b<0$. 
Step 4: Conclusion
$a$ and $b$ must have opposite signs ($a>0, b<0$ or $a<0, b>0$). Option (B) satisfies this. 
Final Answer:(B)