A hyperbola, having the transverse axis of length `2sin theta`, is confocal with the ellipse `3x^2 + 4y^2=12`. Then its equation is

Options

A. (a) `x^(2)"cosec"^(2)theta-y^(2)sec^(2)theta=1`
B. (b) `x^(2)sec^(2)theta-y^(2)"cosec"^(2)theta=1`
C. (c) `x^(2)sin^(2)theta-y^(2)cos^(2)theta=1`
D. (d) `x^(2)cos^(2)theta-y^(2)cos^(2)theta=1`

Explanation

To find the equation of the hyperbola that is confocal with the given ellipse \(3x^2 + 4y^2 = 12\) and has a transverse axis of length \(2 \sin \theta\), we will follow these steps:

### Step 1: Identify the parameters of the ellipse
The given ellipse is \(3x^2 + 4y^2 = 12\). We can rewrite this in standard form by dividing through by 12:

\[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \]

From this, we can identify:
- \(a^2 = 4\) (thus, \(a = 2\))
- \(b^2 = 3\) (thus, \(b = \sqrt{3}\))

### Step 2: Find the value of \(c\) for the ellipse
For an ellipse, the relationship between \(a\), \(b\), and \(c\) (the distance from the center to the foci) is given by:

\[ c^2 = a^2 - b^2 \]

Substituting the values we found:

\[ c^2 = 4 - 3 = 1 \quad \Rightarrow \quad c = 1 \]

### Step 3: Determine the parameters for the hyperbola
Since the hyperbola is confocal with the ellipse, it shares the same value of \(c\). The transverse axis of the hyperbola is given as \(2 \sin \theta\), which means:

\[ 2a' = 2 \sin \theta \quad \Rightarrow \quad a' = \sin \theta \]

### Step 4: Calculate \(b'\) for the hyperbola
For a hyperbola, the relationship between \(a'\), \(b'\), and \(c\) is given by:

\[ c^2 = a'^2 + b'^2 \]

Substituting the known values:

\[ 1^2 = (\sin \theta)^2 + b'^2 \]

This simplifies to:

\[ 1 = \sin^2 \theta + b'^2 \quad \Rightarrow \quad b'^2 = 1 - \sin^2 \theta = \cos^2 \theta \]

Thus, we have:

\[ b' = \cos \theta \]

### Step 5: Write the equation of the hyperbola
The standard form of the equation of a hyperbola centered at the origin with a horizontal transverse axis is:

\[ \frac{x^2}{a'^2} - \frac{y^2}{b'^2} = 1 \]

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

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

### Final Equation
Thus, the equation of the hyperbola is:

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

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