The value of parameter t so that the line `(4-t)x+ty+(a^(3)-1)=0` is normal to the curve xy = 1 may lie in the interval

Options

A. `(1,4)`
B. `(-oo,0)uu(4,oo)`
C. `(-4,4)`
D. `[3,4]`

Explanation

To find the value of the parameter \( t \) such that the line \( (4-t)x + ty + (a^3 - 1) = 0 \) is normal to the curve \( xy = 1 \), we will follow these steps:

### Step 1: Rewrite the line equation in slope-intercept form
The given line equation is:
\[ (4-t)x + ty + (a^3 - 1) = 0 \]
Rearranging this, we get:
\[ ty = -(4-t)x - (a^3 - 1) \]
Dividing through by \( t \) (assuming \( t \neq 0 \)):
\[ y = -\frac{(4-t)}{t}x - \frac{(a^3 - 1)}{t} \]
From this, we can identify the slope \( m \) of the line:
\[ m = -\frac{(4-t)}{t} \]

### Step 2: Find the slope of the normal to the curve
The curve given is \( xy = 1 \). To find the slope of the normal, we first need to find the derivative \( \frac{dy}{dx} \):
1. Rewrite the curve as \( y = \frac{1}{x} \).
2. Differentiate to find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{1}{x^2} \]
3. The slope of the normal line is the negative reciprocal of the slope of the tangent line:
\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx}} = x^2 \]

### Step 3: Set the slopes equal
For the line to be normal to the curve, the slope of the line must equal the slope of the normal:
\[ -\frac{(4-t)}{t} = x^2 \]

### Step 4: Express \( x \) in terms of \( t \)
From the equation \( xy = 1 \), we have \( y = \frac{1}{x} \). Substituting \( y \) into the slope equation:
\[ -\frac{(4-t)}{t} = \left(\frac{1}{y}\right)^2 \]
This implies:
\[ x^2 = -\frac{(4-t)}{t} \]

### Step 5: Solve for \( t \)
Since \( x^2 \) must be non-negative, we need:
\[ -\frac{(4-t)}{t} \geq 0 \]
This leads to two cases:
1. \( 4-t \leq 0 \) and \( t > 0 \) which gives \( t \geq 4 \).
2. \( 4-t \geq 0 \) and \( t < 0 \) which gives \( t < 0 \).

### Final Result
Thus, combining the results, the values of \( t \) that satisfy the condition are:
\[ t \in (-\infty, 0) \cup [4, \infty) \]