The number of values of $c$ such that the line $y = 4x + c$ touches the ellipse $\frac{x^2}{4} + y^2 = 1$ is:
Step-by-Step Solution
Key Concept: Condition for $y = mx + c$ to touch $x^2/a^2 + y^2/b^2 = 1$: $c^2 = a^2m^2 + b^2 = 4(16) + 1 = 65$. So $c = \pm \sqrt{65}$ gives exactly $2$ values.
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Correct Answer: (3)