Question

If the line ax+by+c=0 is a normal to the curve xy=1, then 
 

• A. a>0, b>0
• B. a>0, b<0
• C. a<0,b<0
• D. a=0, b=0

Answer 1

The Correct Option is B

The given curve is \( xy = 1 \), which can be rewritten as:

\[ y = \frac{1}{x} \] This implies that the derivative of \( y \) with respect to \( x \) is: \[ \frac{dy}{dx} = -\frac{1}{x^2} \]

1. Finding the slope of the normal:
The slope of the tangent line to the curve at a given point is the derivative \( \frac{dy}{dx} \). For the normal line, the slope is the negative reciprocal of the tangent slope. Therefore, the slope of the normal is:

\[ m_{\text{normal}} = -\frac{1}{\frac{dy}{dx}} = x^2 \]

2. Equation of the tangent line:
The equation of the tangent line at a point \( (a, \frac{1}{a}) \) on the curve is given by the point-slope form: \[ y - \frac{1}{a} = x^2 (x - a) \] Simplifying this equation: \[ y = a^2 x + \frac{1 - a^3}{a} \] This is the equation of the tangent line to the curve at the point \( (a, \frac{1}{a}) \).

3. Condition for normal line:
For the line to be a normal to the curve at the point \( (a, \frac{1}{a}) \), it must be perpendicular to the curve at that point. The slope of the tangent to the curve at the point \( (a, \frac{1}{a}) \) is given by: \[ \frac{dy}{dx} = -\frac{1}{a^2} \] Therefore, the slope of the line perpendicular to the curve (the normal) is: \[ m_{\text{normal}} = \frac{1}{a^2} \]

4. Product of slopes of tangent and normal:
The tangent and normal lines are perpendicular, so the product of their slopes must be \( -1 \). Hence, the product of the slopes is: \[ x^2 \cdot (-a^2) = -1 \] Solving for \( a \), we get: \[ a = \pm 1 \]

5. Substituting \( a = \pm 1 \) into the equation of the tangent line:
Substituting \( a = 1 \) and \( a = -1 \) into the equation of the tangent line \( y = a^2 x + \frac{1 - a^3}{a} \), we get the two tangent lines:

\[ \text{For } a = 1, \quad y = x + 1 \] \[ \text{For } a = -1, \quad y = -x + 1 \]

6. Conclusion:
The normal at \( (1, 1) \) has a positive slope, and the normal at \( (-1, -1) \) has a negative slope. Therefore, the correct option is:

(B) \( a > 0, b < 0 \)