If the line `x+2y+4=0` cutting the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` in points whose eccentric angies are `30^(@) and 60^(@)` subtends right angle at the origin then its equation is

Options

A. `(x^(2))/(8)+(y^(2))/(4)=1`
B. `(x^(2))/(16)+(y^(2))/(4)=1`
C. `(x^(2))/(4)+(y^(2))/(16)=1`
D. none of these

Answer

To solve the problem, we need to find the equation of the ellipse given that the line \(x + 2y + 4 = 0\) intersects the ellipse at points whose eccentric angles are \(30^\circ\) and \(60^\circ\) and subtends a right angle at the origin.

### Step-by-Step Solution:

1. **Identify the points on the ellipse**:
The points on the ellipse corresponding to the eccentric angles \(30^\circ\) and \(60^\circ\) can be expressed in parametric form as:
\[ P = (a \cos 30^\circ, b \sin 30^\circ) = \left(a \frac{\sqrt{3}}{2}, b \frac{1}{2}\right) \]
\[ Q = (a \cos 60^\circ, b \sin 60^\circ) = \left(a \frac{1}{2}, b \frac{\sqrt{3}}{2}\right) \]

2. **Calculate the coordinates of points P and Q**:
- For point \(P\):
\[ P = \left(a \frac{\sqrt{3}}{2}, b \frac{1}{2}\right) \]
- For point \(Q\):
\[ Q = \left(a \frac{1}{2}, b \frac{\sqrt{3}}{2}\right) \]

3. **Find the slope of line PQ**:
The slope of line segment \(PQ\) is given by:
\[ \text{slope of } PQ = \frac{y_2 - y_1}{x_2 - x_1} = \frac{b \frac{\sqrt{3}}{2} - b \frac{1}{2}}{a \frac{1}{2} - a \frac{\sqrt{3}}{2}} = \frac{b \left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right)}{a \left(\frac{1}{2} - \frac{\sqrt{3}}{2}\right)} \]
Simplifying this gives:
\[ \text{slope of } PQ = \frac{b \left(\frac{\sqrt{3} - 1}{2}\right)}{a \left(\frac{1 - \sqrt{3}}{2}\right)} = -\frac{b(\sqrt{3} - 1)}{a(1 - \sqrt{3})} \]

4. **Find the slope of the line \(x + 2y + 4 = 0\)**:
Rearranging the line equation gives:
\[ 2y = -x - 4 \implies y = -\frac{1}{2}x - 2 \]
The slope of this line is \(-\frac{1}{2}\).

5. **Set the slopes to be negative reciprocals**:
Since the line \(PQ\) subtends a right angle at the origin, we have:
\[ \frac{b(\sqrt{3} - 1)}{a(1 - \sqrt{3})} \cdot \left(-\frac{1}{2}\right) = -1 \]
Simplifying gives:
\[ \frac{b(\sqrt{3} - 1)}{a(1 - \sqrt{3})} = 2 \]

6. **Relate \(a\) and \(b\)**:
From the relationship given, we can express \(b\) in terms of \(a\):
\[ b = \frac{a}{2} \]

7. **Substitute \(b\) into the slope equation**:
Substitute \(b = \frac{a}{2}\) into the slope equation:
\[ \frac{\frac{a}{2}(\sqrt{3} - 1)}{a(1 - \sqrt{3})} = 2 \]
This simplifies to:
\[ \frac{\sqrt{3} - 1}{2(1 - \sqrt{3})} = 2 \]

8. **Solve for \(a\)**:
Cross-multiplying gives:
\[ \sqrt{3} - 1 = 4(1 - \sqrt{3}) \implies \sqrt{3} - 1 = 4 - 4\sqrt{3} \]
Rearranging gives:
\[ 5\sqrt{3} = 5 \implies \sqrt{3} = 1 \implies a = 4 \]

9. **Find \(b\)**:
Since \(b = \frac{a}{2}\):
\[ b = \frac{4}{2} = 2 \]

10. **Write the equation of the ellipse**:
The equation of the ellipse is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \implies \frac{x^2}{16} + \frac{y^2}{4} = 1 \]

### Final Answer:
The equation of the ellipse is:
\[ \frac{x^2}{16} + \frac{y^2}{4} = 1 \]