Step 1: Understanding the Question:
The parametric coordinates of a point on an ellipse are \( (a \cos \phi, b \sin \phi) \), where \( \phi \) is the eccentric angle.
Step 3: Detailed Explanation:
Given ellipse: \( \frac{x^2}{48} + \frac{y^2}{16} = 1 \).
Here, \( a^2 = 48 \implies a = \sqrt{48} = 4\sqrt{3} \).
\( b^2 = 16 \implies b = 4 \).
Point \( P(-6, 2) \) corresponds to:
1. \( x = a \cos \phi = -6 \implies 4\sqrt{3} \cos \phi = -6 \)
\[ \cos \phi = -\frac{6}{4\sqrt{3}} = -\frac{3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2} \]
2. \( y = b \sin \phi = 2 \implies 4 \sin \phi = 2 \)
\[ \sin \phi = \frac{1}{2} \]
Since \( \cos \phi \) is negative and \( \sin \phi \) is positive, \( \phi \) is in the second quadrant.
The reference angle for \( \sin \phi = 1/2 \) is \( 30^\circ \).
In second quadrant: \( \phi = 180^\circ - 30^\circ = 150^\circ \).
Step 4: Final Answer:
The eccentric angle is \( 150^\circ \).