Question

If the line \( a^2 x + ay + 1 = 0 \), for some real number \( a \), is normal to the curve \( xy = 1 \), then:

• A. \( a<0 \)
• B. \( 0<a<1 \)
• C. \( a>0 \)
• D. \( -1<a<1 \)

Hint:

When the line is normal to the curve, use the condition that the product of the slopes of the tangent and normal is -1.

Answer 1

The Correct Option is A

The equation of the curve is \( xy = 1 \). To find the condition when the line is normal to the curve, we need to find the slope of the tangent line at a point on the curve and the slope of the normal line. 
Step 1: Find the slope of the curve. 
Implicitly differentiate the equation of the curve: \[ \frac{d}{dx}(xy) = \frac{d}{dx}(1) \quad \Rightarrow \quad y + x \frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{y}{x}. \] Thus, the slope of the tangent at any point \( (x, y) \) on the curve is \( -\frac{y}{x} \). 
Step 2: Find the slope of the normal.
The slope of the normal line is the negative reciprocal of the slope of the tangent. Therefore, the slope of the normal line is: \[ {Slope of the normal} = \frac{x}{y}. \] Step 3: Find the slope of the given line. 
The equation of the given line is \( a^2 x + ay + 1 = 0 \), which can be rewritten as: \[ y = -\frac{a^2}{a} x - \frac{1}{a} \quad \Rightarrow \quad {Slope of the given line} = -\frac{a^2}{a} = -a. \] Step 4: Condition for the line to be normal to the curve. 
For the line to be normal to the curve, the product of the slopes of the tangent and the normal must be \( -1 \). Thus, we set: \[ \left( \frac{x}{y} \right) \cdot (-a) = -1 \quad \Rightarrow \quad \frac{x}{y} = \frac{1}{a}. \] Since \( xy = 1 \), we substitute \( y = \frac{1}{x} \) into this equation: \[ \frac{x}{\frac{1}{x}} = \frac{1}{a} \quad \Rightarrow \quad x^2 = \frac{1}{a} \quad \Rightarrow \quad a = -1 \quad ({since} \, x^2>0). \] Thus, the correct condition is \( a<0 \). Hence, the correct answer is: \[ \boxed{a<0}. \]