If the latusrectum subtends a right angle at the centre of the hyperbola then its eccentricity

Options

A. `sqrt3`
B. `sqrt3+1`
C. `(sqrt5+1)/2`
D. `(sqrt3+1)/sqrt3`

Answer

To solve the problem step by step, we need to find the eccentricity of a hyperbola given that the latus rectum subtends a right angle at the center of the hyperbola.

### Step 1: Understand the Hyperbola and its Properties
The standard form of a hyperbola centered at the origin is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
The foci of the hyperbola are located at \((ae, 0)\) and \((-ae, 0)\), where \(e\) is the eccentricity defined as:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]

### Step 2: Identify the Coordinates of the Latus Rectum
The latus rectum of the hyperbola is a line segment perpendicular to the transverse axis that passes through the foci. The endpoints of the latus rectum at the focus \((ae, 0)\) are:
\[ \left(ae, \frac{b^2}{a}\right) \quad \text{and} \quad \left(ae, -\frac{b^2}{a}\right) \]
Similarly, for the focus \((-ae, 0)\), the endpoints are:
\[ \left(-ae, \frac{b^2}{a}\right) \quad \text{and} \quad \left(-ae, -\frac{b^2}{a}\right) \]

### Step 3: Analyze the Right Angle Condition
The problem states that the latus rectum subtends a right angle at the center (origin). This means that the angle formed by the lines connecting the center to the endpoints of the latus rectum is \(90^\circ\).

### Step 4: Set Up the Equation
Using the coordinates of the endpoints of the latus rectum, we can form a right triangle with the center. The tangent of the angle formed is given by:
\[ \tan(45^\circ) = 1 = \frac{\text{height}}{\text{base}} = \frac{\frac{b^2}{a}}{ae} \]
This simplifies to:
\[ 1 = \frac{b^2}{a^2 e} \]

### Step 5: Rearranging the Equation
From the equation \(1 = \frac{b^2}{a^2 e}\), we can rearrange it to find:
\[ b^2 = a^2 e \]

### Step 6: Substitute Eccentricity Definition
We know that \(e = \sqrt{1 + \frac{b^2}{a^2}}\). Substituting \(b^2 = a^2 e\) into the eccentricity formula gives:
\[ e = \sqrt{1 + \frac{a^2 e}{a^2}} = \sqrt{1 + e} \]

### Step 7: Square Both Sides
Squaring both sides results in:
\[ e^2 = 1 + e \]
Rearranging gives us:
\[ e^2 - e - 1 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula:
\[ e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]
Since eccentricity cannot be negative, we take the positive root:
\[ e = \frac{1 + \sqrt{5}}{2} \]

### Final Answer
Thus, the eccentricity of the hyperbola is:
\[ e = \frac{1 + \sqrt{5}}{2} \]