The values of parameter 'a' so that the line `(3- a) x+ ay + a^2 -1=0` is a normal to the curve xy = 1 is/are :

Options

A. `(3,oo)`
B. `(-oo,0)`
C. `(0,3)`
D. none

Answer

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

### Step 1: Find the slope of the normal to the curve
The given curve is \(xy = 1\). To find the slope of the normal, we first need to find the derivative of the curve.

Starting from the equation of the curve:
\[ xy = 1 \]

Differentiating both sides with respect to \(x\):
\[ x \frac{dy}{dx} + y = 0 \]

From this, we can express \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = -\frac{y}{x} \]

### Step 2: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent line:
\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx}} = \frac{x}{y} \]

### Step 3: Express \(y\) in terms of \(x\)
From the curve \(xy = 1\), we can express \(y\) as:
\[ y = \frac{1}{x} \]

### Step 4: Substitute \(y\) into the slope of the normal
Substituting \(y\) into the slope of the normal:
\[ \text{slope of normal} = \frac{x}{\frac{1}{x}} = x^2 \]

### Step 5: Find the slope of the given line
Now, we need to rewrite the line equation \((3 - a)x + ay + a^2 - 1 = 0\) in slope-intercept form \(y = mx + c\).

Rearranging the line equation:
\[ ay = -(3 - a)x + 1 - a^2 \]
\[ y = -\frac{(3 - a)}{a}x + \frac{1 - a^2}{a} \]

From this, we can identify the slope \(m\) of the line:
\[ m = -\frac{(3 - a)}{a} \]

### Step 6: Set the slopes equal
For the line to be a normal to the curve, the slope of the normal must equal the slope of the line:
\[ x^2 = -\frac{(3 - a)}{a} \]

### Step 7: Analyze the conditions
Since \(x^2 \geq 0\), we need:
\[ -\frac{(3 - a)}{a} \geq 0 \]

This inequality implies:
1. \(3 - a \leq 0\) and \(a > 0\) (both must be true)
2. or \(3 - a \geq 0\) and \(a < 0\) (both must be true)

### Step 8: Solve the inequalities
From \(3 - a \leq 0\):
\[ a \geq 3 \]

From \(3 - a \geq 0\):
\[ a \leq 3 \]

Thus, we have:
- \(a < 0\) gives us \(a \in (-\infty, 0)\)
- \(a \geq 3\) gives us \(a \in [3, \infty)\)

### Final Answer
The values of the parameter \(a\) such that the line is a normal to the curve are:
\[ a \in (-\infty, 0) \cup [3, \infty) \]