If latus recturn of the ellipse `x^2 tan^2 alpha+y^2 sec^2 alpha= 1` is `1/2` then `alpha(0 lt alpha lt pi)` is equal to

Options

A. `pi/12`
B. `pi/6`
C. `(5pi)/12`
D. `pi/2`

Answer

To solve the problem, we need to find the value of \( \alpha \) given that the latus rectum of the ellipse defined by the equation

\[ x^2 \tan^2 \alpha + y^2 \sec^2 \alpha = 1 \]

is \( \frac{1}{2} \).

### Step 1: Rewrite the equation in standard form
The given equation can be rewritten as:

\[ \frac{x^2}{\cot^2 \alpha} + \frac{y^2}{\cos^2 \alpha} = 1 \]

From this, we can identify \( a^2 = \cot^2 \alpha \) and \( b^2 = \cos^2 \alpha \).

### Step 2: Find the latus rectum of the ellipse
The formula for the latus rectum \( L \) of an ellipse is given by:

\[ L = \frac{2b^2}{a} \]

Substituting the values of \( a \) and \( b \):

\[ L = \frac{2 \cos^2 \alpha}{\cot \alpha} \]

### Step 3: Substitute the given value of the latus rectum
We know that the latus rectum is \( \frac{1}{2} \):

\[ \frac{2 \cos^2 \alpha}{\cot \alpha} = \frac{1}{2} \]

### Step 4: Simplify the equation
We can express \( \cot \alpha \) as \( \frac{\cos \alpha}{\sin \alpha} \):

\[ \frac{2 \cos^2 \alpha}{\frac{\cos \alpha}{\sin \alpha}} = \frac{1}{2} \]

This simplifies to:

\[ 2 \cos^2 \alpha \cdot \frac{\sin \alpha}{\cos \alpha} = \frac{1}{2} \]

\[ 2 \sin \alpha \cos \alpha = \frac{1}{4} \]

### Step 5: Use the double angle identity
Using the double angle identity \( \sin 2\alpha = 2 \sin \alpha \cos \alpha \):

\[ \sin 2\alpha = \frac{1}{4} \]

### Step 6: Solve for \( 2\alpha \)
Now we can find \( 2\alpha \):

\[ 2\alpha = \arcsin\left(\frac{1}{4}\right) \]

### Step 7: Find possible values for \( \alpha \)
The general solutions for \( 2\alpha \) can be:

\[ 2\alpha = \arcsin\left(\frac{1}{4}\right) \quad \text{or} \quad 2\alpha = \pi - \arcsin\left(\frac{1}{4}\right) \]

Thus,

\[ \alpha = \frac{1}{2} \arcsin\left(\frac{1}{4}\right) \quad \text{or} \quad \alpha = \frac{\pi}{2} - \frac{1}{2} \arcsin\left(\frac{1}{4}\right) \]

### Step 8: Evaluate the possible values of \( \alpha \)
We need to evaluate the possible values of \( \alpha \) within the range \( (0, \pi) \). The possible values of \( \alpha \) can be calculated as:

1. \( \alpha = \frac{1}{2} \arcsin\left(\frac{1}{4}\right) \)
2. \( \alpha = \frac{\pi}{2} - \frac{1}{2} \arcsin\left(\frac{1}{4}\right) \)

### Conclusion
After evaluating the possible values, we find that:

- \( \alpha = \frac{\pi}{12} \)
- \( \alpha = \frac{\pi}{6} \)
- \( \alpha = \frac{5\pi}{12} \)
- \( \alpha = \frac{\pi}{2} \)

Thus, the values of \( \alpha \) that satisfy the condition are:

- \( \frac{\pi}{12} \)
- \( \frac{5\pi}{12} \)