Question

The eccentric angle of the point ( P(-6, 2) ) of the ellipse ( \frac{x^2}{48} + \frac{y^2}{16} = 1 ) is

• A. ( 30^\circ )
• B. ( 135^\circ )
• C. ( 150^\circ )
• D. [suspicious link removed]

Hint:

Eccentric angle $\phi$ is measured from the major axis in a counter-clockwise direction.

Answer 1

The Correct Option is C

Step 1: Identify Parameters
(a^2 = 48 \implies a = 4\sqrt{3}).
(b^2 = 16 \implies b = 4).

Step 2: Coordinate Comparison
Point (P) on ellipse is ((a \cos \phi, b \sin \phi)).
(4\sqrt{3} \cos \phi = -6 \implies \cos \phi = -\frac{6}{4\sqrt{3}} = -\frac{\sqrt{3}}{2}).
(4 \sin \phi = 2 \implies \sin \phi = \frac{1}{2}).

Step 3: Determine Angle
(\cos \phi = -\frac{\sqrt{3}}{2}) and (\sin \phi = \frac{1}{2}) place the point in the second quadrant.
(\phi = 180^\circ - 30^\circ = 150^\circ).
Final Answer: (C)