If the tangent at the point `P(a sec theta , b tan theta)` to the hyperbola passes through the point where a directrix of the hyperbola meets the positive side of the transverse axis, then `theta` is equal to

Options

A. `cos^(-1)(1//e)`
B. `tan^(-1)(1//e)`
C. `cot^(-1)(1//e)`
D. `sec^(-1)(1//e)`

Answer

To solve the problem, we need to find the value of \(\theta\) given that the tangent at the point \(P(a \sec \theta, b \tan \theta)\) on the hyperbola passes through the point where the directrix meets the positive side of the transverse axis.

### Step-by-Step Solution:

1. **Identify the Hyperbola**: The standard form of the hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

2. **Directrix of the Hyperbola**: The directrix of the hyperbola is given by the equations:
\[ x = \frac{a}{e} \quad \text{and} \quad x = -\frac{a}{e} \]
where \(e\) is the eccentricity of the hyperbola, defined as \(e = \sqrt{1 + \frac{b^2}{a^2}}\).

3. **Point of Tangency**: The point \(P\) on the hyperbola is given as:
\[ P\left(a \sec \theta, b \tan \theta\right) \]

4. **Equation of the Tangent Line**: The equation of the tangent to the hyperbola at the point \(P(x_1, y_1)\) is given by:
\[ \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \]
Substituting \(x_1 = a \sec \theta\) and \(y_1 = b \tan \theta\), we have:
\[ \frac{x(a \sec \theta)}{a^2} - \frac{y(b \tan \theta)}{b^2} = 1 \]
Simplifying this, we get:
\[ \frac{x \sec \theta}{a} - \frac{y \tan \theta}{b} = 1 \]

5. **Substituting the Point of Intersection**: The tangent line passes through the point where the directrix meets the positive side of the transverse axis, which is \(\left(\frac{a}{e}, 0\right)\). Substituting \(x = \frac{a}{e}\) and \(y = 0\) into the tangent equation:
\[ \frac{\frac{a}{e} \sec \theta}{a} - \frac{0 \cdot \tan \theta}{b} = 1 \]
This simplifies to:
\[ \frac{\sec \theta}{e} = 1 \]

6. **Finding \(\theta\)**: From the equation \(\frac{\sec \theta}{e} = 1\), we can express \(\sec \theta\) as:
\[ \sec \theta = e \]
Therefore, taking the reciprocal, we have:
\[ \cos \theta = \frac{1}{e} \]

7. **Final Result**: Since \(e = \sqrt{1 + \frac{b^2}{a^2}}\), we can conclude that:
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{1 + \frac{b^2}{a^2}}}\right) \]