The value of the parameter `a` so that the line `(3-a)x+ay+ a^2-1=0` is normal to the curve `xy=1` may lie in the interval

Options

A. `(-oo, 0)cup (3,oo)`
B. (1,3)
C. (-3,3)
D. None of these

Answer

To find the value of the parameter \( a \) such that the line \( (3-a)x + ay + a^2 - 1 = 0 \) is normal to the curve \( xy = 1 \), we will follow these steps:

### Step 1: Rewrite the curve
The curve \( xy = 1 \) can be rewritten as:
\[ y = \frac{1}{x} \]

### Step 2: Differentiate the curve
To find the slope of the tangent to the curve, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = -\frac{1}{x^2} \]

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Thus, the slope of the normal line is:
\[ \text{slope of normal} = -\left(-\frac{1}{x^2}\right)^{-1} = x^2 \]

### Step 4: Find the slope of the given line
The given line can be rearranged to find its slope. The line equation is:
\[ (3-a)x + ay + a^2 - 1 = 0 \]
Rearranging gives:
\[ ay = -(3-a)x + 1 - a^2 \]
Thus, the slope of the line is:
\[ \text{slope of line} = -\frac{3-a}{a} \]

### Step 5: Set the slopes equal
For the line to be normal to the curve, the slope of the normal must equal the slope of the line:
\[ x^2 = -\frac{3-a}{a} \]

### Step 6: Determine the conditions for \( a \)
Since \( x^2 \) is always non-negative, we require:
\[ -\frac{3-a}{a} \geq 0 \]
This inequality implies two cases:
1. \( 3-a \leq 0 \) and \( a > 0 \)
2. \( 3-a \geq 0 \) and \( a < 0 \)

### Step 7: Solve the inequalities
1. From \( 3-a \leq 0 \):
\[ a \geq 3 \]
2. From \( 3-a \geq 0 \):
\[ a \leq 3 \quad \text{and} \quad a < 0 \Rightarrow a < 0 \]

### Step 8: Combine the results
From the analysis:
- \( a \geq 3 \) gives us values in the interval \( [3, \infty) \).
- \( a < 0 \) gives us values in the interval \( (-\infty, 0) \).

Thus, the values of \( a \) that satisfy the condition are:
\[ a \in (-\infty, 0) \cup [3, \infty) \]

### Final Answer
The value of the parameter \( a \) so that the line is normal to the curve \( xy = 1 \) may lie in the intervals \( (-\infty, 0) \) or \( [3, \infty) \).