Step-by-step Solution:

Step 1: General Condition for Tangency
For a line \( y = mx + c \) to touch the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the condition is: \[ c^2 = a^2 m^2 + b^2 \] Step 2: Identify Given Parameters
The given equation of the ellipse is: \[ x^2 + 4y^2 = 4 \] Rewriting it in standard form: \[ \frac{x^2}{4} + \frac{y^2}{1} = 1 \] Comparing with the standard equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we get: \[ a^2 = 4, \quad b^2 = 1 \] The given line is: \[ y = 4x + c \] So the slope is \( m = 4 \). Step 3: Apply Tangency Condition
Using the tangency condition: \[ c^2 = a^2 m^2 + b^2 \] Substituting the values: \[ c^2 = (4)(16) + 1 \] \[ c^2 = 64 + 1 = 65 \] \[ c = \pm \sqrt{65} \] Step 4: Conclusion Since there are two values of \( c \) (i.e., \( \sqrt{65} \) and \( -\sqrt{65} \)), the correct answer is 2. Thus, the correct option is A (2).