P is the extremity of the latuscrectum of ellipse `3x^(2)+4y^(2)=48` in the first quadrant. The eccentric angle of P is

Options

A. `pi/8`
B. `(3pi)/(4)`
C. `pi/3`
D. `(2pi)/(3)`

Answer

To find the eccentric angle of point P, which is the extremity of the latus rectum of the ellipse given by the equation \(3x^2 + 4y^2 = 48\), we will follow these steps:

### Step 1: Convert the equation of the ellipse to standard form
The standard form of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Starting with the given equation:
\[ 3x^2 + 4y^2 = 48 \]
We can divide the entire equation by 48:
\[ \frac{3x^2}{48} + \frac{4y^2}{48} = 1 \]
This simplifies to:
\[ \frac{x^2}{16} + \frac{y^2}{12} = 1 \]
Thus, we have \(a^2 = 16\) and \(b^2 = 12\).

### Step 2: Identify the values of \(a\) and \(b\)
From the above, we find:
\[ a = \sqrt{16} = 4 \quad \text{and} \quad b = \sqrt{12} = 2\sqrt{3} \]

### Step 3: Calculate the eccentricity \(e\)
The eccentricity \(e\) of the ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values of \(b^2\) and \(a^2\):
\[ e = \sqrt{1 - \frac{12}{16}} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Step 4: Find the coordinates of point P
The coordinates of the extremity of the latus rectum \(P\) can be found using the formulas:
\[ P = (ae, \frac{b^2}{a}) \]
Substituting the values of \(a\), \(b\), and \(e\):
\[ P = \left(4 \cdot \frac{1}{2}, \frac{12}{4}\right) = (2, 3) \]

### Step 5: Calculate the eccentric angle \(\theta\)
The eccentric angle \(\theta\) is given by the formulas:
\[ x = a \cos \theta \quad \text{and} \quad y = b \sin \theta \]
Substituting the coordinates of point \(P\):
\[ 2 = 4 \cos \theta \quad \text{and} \quad 3 = 2\sqrt{3} \sin \theta \]
From the first equation:
\[ \cos \theta = \frac{2}{4} = \frac{1}{2} \]
This implies:
\[ \theta = \frac{\pi}{3} \quad \text{(since \(\cos \frac{\pi}{3} = \frac{1}{2}\))} \]

### Conclusion
Thus, the eccentric angle of point \(P\) is:
\[ \theta = \frac{\pi}{3} \]