The line `y = 4x + c ` touches the hyperbola `x^(2) - y^(2) = 1` if

Options

A. `c = 0`
B. `c = pm sqrt 15`
C. `c = pm sqrt2`
D. none of these

Answer

To determine the value of \( c \) for which the line \( y = 4x + c \) touches the hyperbola \( x^2 - y^2 = 1 \), we can follow these steps:

### Step 1: Identify the hyperbola and its parameters
The given hyperbola is \( x^2 - y^2 = 1 \). This is a standard form of a hyperbola centered at the origin with \( a^2 = 1 \) and \( b^2 = 1 \).

### Step 2: Write the equation of the tangent line
The equation of the line is given as \( y = 4x + c \). Here, the slope \( m \) of the line is 4.

### Step 3: Use the condition for tangency
For the line to be tangent to the hyperbola, the distance from the center of the hyperbola to the line must equal the distance to the hyperbola along the direction of the slope. The condition for tangency can be derived from the formula:
\[ c = \pm \sqrt{a^2 m^2 - b^2} \]
where \( a^2 = 1 \), \( b^2 = 1 \), and \( m = 4 \).

### Step 4: Substitute the values into the tangency condition
Substituting the values into the formula gives:
\[ c = \pm \sqrt{1 \cdot 4^2 - 1} \]
\[ c = \pm \sqrt{16 - 1} \]
\[ c = \pm \sqrt{15} \]

### Step 5: Conclusion
Thus, the line \( y = 4x + c \) touches the hyperbola \( x^2 - y^2 = 1 \) if:
\[ c = \sqrt{15} \quad \text{or} \quad c = -\sqrt{15} \]