If the tangent at the point `(asec alpha, b tanalpha )` to the hyberbola `(x^(2))/(a^(2)) -(y^(2))/(b^(2)) =1` meets the transverse axis at T. Then the distances of T form a focus of the hyperbola is

Options

A. `a(e -cos alpha)`
B. `b( e + cos alpha)`
C. `a(e + cos alpha)`
D. `sqrt(a^(2)e^(2) + b^(2) cot^(2) alpha)`

Answer

To solve the problem, we need to find the distance of the point \( T \) (where the tangent at the point \( (a \sec \alpha, b \tan \alpha) \) meets the transverse axis of the hyperbola) from one of the foci of the hyperbola defined by the equation:

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

### Step 1: Identify the point on the hyperbola
The point given is \( P(a \sec \alpha, b \tan \alpha) \).

### Step 2: Write the equation of the tangent line at point \( P \)
The equation of the tangent to the hyperbola at point \( P(x_1, y_1) \) can be expressed as:

\[ \frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1 \]

Substituting \( x_1 = a \sec \alpha \) and \( y_1 = b \tan \alpha \):

\[ \frac{x (a \sec \alpha)}{a^2} - \frac{y (b \tan \alpha)}{b^2} = 1 \]

This simplifies to:

\[ \frac{x \sec \alpha}{a} - \frac{y \tan \alpha}{b} = 1 \]

### Step 3: Find the intersection of the tangent with the transverse axis
The transverse axis corresponds to \( y = 0 \). Substituting \( y = 0 \) into the tangent equation:

\[ \frac{x \sec \alpha}{a} = 1 \]

Solving for \( x \):

\[ x = a \cos \alpha \]

Thus, the coordinates of point \( T \) are \( (a \cos \alpha, 0) \).

### Step 4: Find the distance from point \( T \) to the focus of the hyperbola
The foci of the hyperbola are located at \( (ae, 0) \) and \( (-ae, 0) \), where \( e = \sqrt{1 + \frac{b^2}{a^2}} \).

The distance \( d \) from point \( T(a \cos \alpha, 0) \) to the focus \( (ae, 0) \) is given by:

\[ d = |a \cos \alpha - ae| \]

### Step 5: Simplify the distance expression
Factoring out \( a \):

\[ d = a | \cos \alpha - e | \]

### Final Result
Thus, the distance of point \( T \) from the focus of the hyperbola is:

\[ d = a | e - \cos \alpha | \]