If the tangent at the point `(2sec theta,3tan theta)` to the hyperbola `(x^(2))/(4)-(y^(2))/(9)=1` is parallel to `3x-4y+4=0`, then the value of `theta`, is

Options

A. `(pi)/(6)`
B. `(pi)/(4)`
C. `(pi)/(3)`
D. `(5pi)/(12)`

Answer

To solve the problem step by step, we will follow the logic presented in the video transcript.

### Step 1: Identify the hyperbola and the point
The given hyperbola is:
\[ \frac{x^2}{4} - \frac{y^2}{9} = 1 \]
The point on the hyperbola is given as:
\[ (2 \sec \theta, 3 \tan \theta) \]

### Step 2: Write the equation of the tangent line
The equation of the tangent to the hyperbola at the point \((x_0, y_0)\) is given by:
\[ \frac{x_0 x}{a^2} - \frac{y_0 y}{b^2} = 1 \]
where \(a^2 = 4\) and \(b^2 = 9\). Substituting \(x_0 = 2 \sec \theta\) and \(y_0 = 3 \tan \theta\):
\[ \frac{(2 \sec \theta) x}{4} - \frac{(3 \tan \theta) y}{9} = 1 \]
This simplifies to:
\[ \frac{2 \sec \theta}{4} x - \frac{3 \tan \theta}{9} y = 1 \]
or:
\[ \frac{1}{2} \sec \theta x - \frac{1}{3} \tan \theta y = 1 \]

### Step 3: Determine the slope of the tangent line
The slope of the line in the form \(Ax + By + C = 0\) is given by \(-\frac{A}{B}\). Here, we rearrange the tangent equation:
\[ \frac{1}{2} \sec \theta x - \frac{1}{3} \tan \theta y - 1 = 0 \]
Thus, \(A = \frac{1}{2} \sec \theta\) and \(B = -\frac{1}{3} \tan \theta\). The slope \(m\) of the tangent line is:
\[ m = -\frac{\frac{1}{2} \sec \theta}{-\frac{1}{3} \tan \theta} = \frac{\frac{1}{2} \sec \theta}{\frac{1}{3} \tan \theta} = \frac{3 \sec \theta}{2 \tan \theta} \]

### Step 4: Find the slope of the given line
The given line is:
\[ 3x - 4y + 4 = 0 \]
The slope of this line is:
\[ m' = -\frac{3}{-4} = \frac{3}{4} \]

### Step 5: Set the slopes equal
Since the tangent line is parallel to the given line, we set the slopes equal:
\[ \frac{3 \sec \theta}{2 \tan \theta} = \frac{3}{4} \]

### Step 6: Simplify the equation
Cross-multiplying gives:
\[ 3 \sec \theta \cdot 4 = 3 \cdot 2 \tan \theta \]
This simplifies to:
\[ 12 \sec \theta = 6 \tan \theta \]
Dividing both sides by 6:
\[ 2 \sec \theta = \tan \theta \]

### Step 7: Use trigonometric identities
Recall that:
\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Substituting these into the equation gives:
\[ 2 \cdot \frac{1}{\cos \theta} = \frac{\sin \theta}{\cos \theta} \]
Multiplying both sides by \(\cos \theta\) (assuming \(\cos \theta \neq 0\)):
\[ 2 = \sin \theta \]

### Step 8: Solve for \(\theta\)
Since \(\sin \theta = \frac{1}{2}\), we find:
\[ \theta = 30^\circ \quad \text{(or } \theta = 150^\circ \text{, but only } 30^\circ \text{ is valid in the context)} \]

### Final Answer
The value of \(\theta\) is:
\[ \theta = 30^\circ \]