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 \( c \) such that the line \( y = 4x + c \) touches the hyperbola \( x^2 - y^2 = 1 \), we can follow these steps:

### Step 1: Understand the Condition for Tangency
A line touches a curve if there is exactly one point of intersection between them. This means that when we substitute the equation of the line into the equation of the hyperbola, the resulting quadratic equation should have exactly one solution (i.e., the discriminant should be zero).

### Step 2: Substitute the Line Equation into the Hyperbola Equation
We have the hyperbola equation:
\[ x^2 - y^2 = 1 \]
Substituting \( y = 4x + c \) into the hyperbola equation:
\[ x^2 - (4x + c)^2 = 1 \]

### Step 3: Expand the Equation
Expanding the equation:
\[ x^2 - (16x^2 + 8cx + c^2) = 1 \]
This simplifies to:
\[ x^2 - 16x^2 - 8cx - c^2 = 1 \]
\[ -15x^2 - 8cx - c^2 - 1 = 0 \]
Multiplying through by -1 gives:
\[ 15x^2 + 8cx + (c^2 + 1) = 0 \]

### Step 4: Find the Discriminant
For the quadratic equation \( Ax^2 + Bx + C = 0 \), the discriminant \( D \) is given by:
\[ D = B^2 - 4AC \]
Here, \( A = 15 \), \( B = 8c \), and \( C = c^2 + 1 \). Thus, the discriminant is:
\[ D = (8c)^2 - 4 \cdot 15 \cdot (c^2 + 1) \]
\[ D = 64c^2 - 60(c^2 + 1) \]
\[ D = 64c^2 - 60c^2 - 60 \]
\[ D = 4c^2 - 60 \]

### Step 5: Set the Discriminant to Zero
For the line to be a tangent to the hyperbola, we set the discriminant equal to zero:
\[ 4c^2 - 60 = 0 \]
\[ 4c^2 = 60 \]
\[ c^2 = 15 \]
\[ c = \pm \sqrt{15} \]

### Conclusion
The values of \( c \) for which the line \( y = 4x + c \) touches the hyperbola \( x^2 - y^2 = 1 \) are:
\[ c = \sqrt{15} \quad \text{and} \quad c = -\sqrt{15} \]