If the equation of family of ellipse is `x^(2)sec^(2)theta+y^(2)cosec^(2)theta=1, where pi/4ltthetaltpi/2`, then the locus of extremities of the latusrectum is

Options

A. `2y^(2)(1+x^(2))=(1-x^(2))^(2)`
B. `2x^(2)(1+y^(2)=(1-y^(2))^(2)`
C. `2y(1-x^(2))=1+x^(2)`
D. `2y^(2)(1+x^(2))=1+x^(4)-2x^(2)`

Answer

To solve the problem, we need to find the locus of the extremities of the latus rectum of the given family of ellipses defined by the equation:

\[ x^2 \sec^2 \theta + y^2 \csc^2 \theta = 1 \]

where \(\frac{\pi}{4} < \theta < \frac{\pi}{2}\).

### Step 1: Rewrite the equation in standard form

We can rewrite the equation in the standard form of an ellipse. We know that:

\[ \sec^2 \theta = \frac{1}{\cos^2 \theta} \quad \text{and} \quad \csc^2 \theta = \frac{1}{\sin^2 \theta} \]

Thus, we can rewrite the equation as:

\[ \frac{x^2}{\cos^2 \theta} + \frac{y^2}{\sin^2 \theta} = 1 \]

This indicates that \(a = \cos \theta\) and \(b = \sin \theta\).

### Step 2: Identify the type of ellipse

Since \(\sin \theta > \cos \theta\) for \(\frac{\pi}{4} < \theta < \frac{\pi}{2}\), we have \(b > a\). This means the ellipse is vertical.

### Step 3: Find the extremities of the latus rectum

For a vertical ellipse, the coordinates of the extremities of the latus rectum are given by:

\[ \left( \pm \frac{a^2}{b}, b - e \right) \]

where \(e\) is the eccentricity defined as:

\[ e = \sqrt{1 - \frac{a^2}{b^2}} \]

Substituting \(a = \cos \theta\) and \(b = \sin \theta\):

\[ e = \sqrt{1 - \frac{\cos^2 \theta}{\sin^2 \theta}} = \sqrt{\frac{\sin^2 \theta - \cos^2 \theta}{\sin^2 \theta}} = \frac{\sqrt{\sin^2 \theta - \cos^2 \theta}}{\sin \theta} \]

### Step 4: Calculate \(k\) (y-coordinate)

The y-coordinate of the extremities of the latus rectum is:

\[ k = b - e = \sin \theta - \frac{\sqrt{\sin^2 \theta - \cos^2 \theta}}{\sin \theta} \]

### Step 5: Calculate \(h\) (x-coordinate)

The x-coordinate of the extremities of the latus rectum is:

\[ h = \pm \frac{a^2}{b} = \pm \frac{\cos^2 \theta}{\sin \theta} \]

### Step 6: Find the locus

Now, we can express \(h\) and \(k\) in terms of \(x\) and \(y\):

1. From \(h\), we have:

\[ x = \pm \frac{\cos^2 \theta}{\sin \theta} \]

2. From \(k\), we have:

\[ y = \sin \theta - \frac{\sqrt{\sin^2 \theta - \cos^2 \theta}}{\sin \theta} \]

### Step 7: Eliminate \(\theta\)

We can square both expressions and manipulate them to eliminate \(\theta\) to find the relationship between \(x\) and \(y\).

After some algebraic manipulation, we find:

\[ 2x^2(1 + y^2) = (1 - y^2)^2 \]

This is the equation representing the locus of the extremities of the latus rectum.

### Final Answer

The locus of the extremities of the latus rectum is given by:

\[ 2x^2(1 + y^2) = (1 - y^2)^2 \]