If P and Q are points with eccentric angles `theta` and `(theta+(pi)/(6))` on the ellipse `(x^(2))/(16)+(y^(2))/(4)=1`, then the area (in sq. units) of the triangle OPQ (where O is the origin) is equal to

Answer

To find the area of triangle OPQ where O is the origin and P and Q are points on the ellipse given by the equation \(\frac{x^2}{16} + \frac{y^2}{4} = 1\) with eccentric angles \(\theta\) and \(\theta + \frac{\pi}{6}\), we will follow these steps:

### Step 1: Identify the coordinates of points P and Q
The coordinates of a point on the ellipse can be expressed as:
\[ (x, y) = (a \cos \theta, b \sin \theta) \]
where \(a\) and \(b\) are the semi-major and semi-minor axes of the ellipse, respectively. For the given ellipse:
- \(a = 4\) (since \(16 = 4^2\))
- \(b = 2\) (since \(4 = 2^2\))

Thus, the coordinates of point P (with eccentric angle \(\theta\)) are:
\[ P = (4 \cos \theta, 2 \sin \theta) \]

For point Q (with eccentric angle \(\theta + \frac{\pi}{6}\)):
\[ Q = (4 \cos(\theta + \frac{\pi}{6}), 2 \sin(\theta + \frac{\pi}{6})) \]

### Step 2: Calculate the coordinates of point Q
Using the angle addition formulas:
\[ \cos(\theta + \frac{\pi}{6}) = \cos \theta \cos \frac{\pi}{6} - \sin \theta \sin \frac{\pi}{6} = \cos \theta \cdot \frac{\sqrt{3}}{2} - \sin \theta \cdot \frac{1}{2} \]
\[ \sin(\theta + \frac{\pi}{6}) = \sin \theta \cos \frac{\pi}{6} + \cos \theta \sin \frac{\pi}{6} = \sin \theta \cdot \frac{\sqrt{3}}{2} + \cos \theta \cdot \frac{1}{2} \]

Thus, the coordinates of point Q are:
\[ Q = \left(4 \left(\cos \theta \cdot \frac{\sqrt{3}}{2} - \sin \theta \cdot \frac{1}{2}\right), 2 \left(\sin \theta \cdot \frac{\sqrt{3}}{2} + \cos \theta \cdot \frac{1}{2}\right)\right) \]

### Step 3: Use the area formula for triangle OPQ
The area \(A\) of triangle OPQ can be calculated using the determinant formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
- \(O(0, 0)\)
- \(P(4 \cos \theta, 2 \sin \theta)\)
- \(Q\left(4 \left(\cos \theta \cdot \frac{\sqrt{3}}{2} - \sin \theta \cdot \frac{1}{2}\right), 2 \left(\sin \theta \cdot \frac{\sqrt{3}}{2} + \cos \theta \cdot \frac{1}{2}\right)\right)\)

This gives:
\[ A = \frac{1}{2} \left| 0 + 4 \cos \theta \left(2 \left(\sin \theta \cdot \frac{\sqrt{3}}{2} + \cos \theta \cdot \frac{1}{2}\right) - 2 \sin \theta\right) \right| \]
\[ = \frac{1}{2} \left| 4 \cos \theta \left(2 \sin \theta \cdot \frac{\sqrt{3}}{2} + 2 \cos \theta \cdot \frac{1}{2} - 2 \sin \theta\right) \right| \]
\[ = 2 \left| \cos \theta \left(\sqrt{3} \sin \theta + \cos \theta - \sin \theta\right) \right| \]
\[ = 2 \left| \cos \theta \left((\sqrt{3} - 1) \sin \theta + \cos \theta\right) \right| \]

### Step 4: Final Area Calculation
The final area of triangle OPQ is:
\[ A = 2 \left| \cos \theta \left((\sqrt{3} - 1) \sin \theta + \cos \theta\right) \right| \]