If normal to hyperbola `x^2/a^2-y^2/b^2=1` drawn at an extremity of its latus-rectum has slope equal to the slope of line which meets hyperbola only once, then the eccentricity of hyperbola is

Options

A. `e = sqrt((1+sqrt(5))/(2))`
B. `e = sqrt((sqrt(5)+3)/(2))`
C. `e = sqrt((2)/(sqrt(5)-1))`
D. None of these

Explanation

To solve the problem step by step, we will analyze the given hyperbola and the conditions provided in the question.

### Step 1: Identify the extremities of the latus rectum of the hyperbola
The equation of the hyperbola is given as:

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

The extremities of the latus rectum for this hyperbola are located at the points:

\[ \left( ae, \frac{b^2}{a} \right) \quad \text{and} \quad \left( -ae, \frac{b^2}{a} \right) \]

where \( e \) is the eccentricity of the hyperbola.

### Step 2: Write the equation of the normal at the extremity of the latus rectum
The equation of the normal to the hyperbola at a point \( (x_1, y_1) \) is given by:

\[ \frac{a^2 y_1}{b^2} (x - x_1) + b^2 (y - y_1) = 0 \]

Substituting \( x_1 = ae \) and \( y_1 = \frac{b^2}{a} \):

\[ \frac{a^2 \cdot \frac{b^2}{a}}{b^2} (x - ae) + b^2 \left( y - \frac{b^2}{a} \right) = 0 \]

This simplifies to:

\[ a (x - ae) + b^2 \left( y - \frac{b^2}{a} \right) = 0 \]

### Step 3: Rearranging the equation of the normal
Rearranging gives:

\[ a x - a^2 e + b^2 y - \frac{b^4}{a} = 0 \]

This can be expressed in slope-intercept form:

\[ b^2 y = -a x + a^2 e + \frac{b^4}{a} \]

Thus, the slope \( m \) of the normal line is:

\[ m = -\frac{a}{b^2} \]

### Step 4: Condition for the line meeting the hyperbola only once
The slope of a line that meets the hyperbola only once is equal to the slope of the tangent to the hyperbola at the point of tangency. The slope of the tangent line at a point on the hyperbola can be expressed as:

\[ m_t = \frac{b \sec \theta}{a \tan \theta} \]

Setting the slopes equal gives:

\[ -\frac{1}{e} = \frac{b \sec \theta}{a \tan \theta} \]

### Step 5: Relating slopes and eccentricity
From the above equation, we can manipulate it to find a relationship involving \( e \):

\[ \sec \theta = \frac{1}{\cos \theta}, \quad \tan \theta = \frac{\sin \theta}{\cos \theta} \]

Thus, we can express:

\[ -\frac{1}{e} = \frac{b}{a} \cdot \frac{1}{\sin \theta} \]

Cross-multiplying gives:

\[ \sin \theta = -\frac{b}{a e} \]

### Step 6: Using the identity for eccentricity
Using the identity \( e^2 = 1 + \frac{b^2}{a^2} \), we can substitute \( \frac{b^2}{a^2} \) in terms of \( e \):

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

### Step 7: Solving for \( e \)
Cross-multiplying and simplifying will lead to a polynomial in \( e^2 \):

\[ e^4 - e^2 - 1 = 0 \]

Using the quadratic formula:

\[ e^2 = \frac{1 \pm \sqrt{5}}{2} \]

Since \( e \) must be positive, we take the positive root:

\[ e = \sqrt{\frac{1 + \sqrt{5}}{2}} \]

### Final Answer
Thus, the eccentricity of the hyperbola is:

\[ e = \sqrt{\frac{1 + \sqrt{5}}{2}} \]