The eccentricity of the hyperola whose latus rectum subtends a right angle at centre is

Options

A. `2 sin 18^(@)`
B. `2 cos 36^(@)`
C. `2 tan 18^@`
D. `2 cot 18^@`

Answer

To find the eccentricity of the hyperbola whose latus rectum subtends a right angle at the center, we can follow these steps:

### Step 1: Understand the Hyperbola and Latus Rectum
The latus rectum of a hyperbola is a line segment perpendicular to the transverse axis, passing through a focus and whose endpoints lie on the hyperbola. For a hyperbola centered at the origin, the coordinates of the focus are \((ae, 0)\) and the endpoints of the latus rectum are \((ae, \frac{b^2}{a})\) and \((ae, -\frac{b^2}{a})\).

### Step 2: Set Up the Right Angle Condition
The problem states that the latus rectum subtends a right angle at the center of the hyperbola (the origin). This means that the angle formed by the lines connecting the origin to the endpoints of the latus rectum is \(90^\circ\).

### Step 3: Use Tangent of the Angle
Since the angle is \(90^\circ\), we can use the tangent of \(45^\circ\) (which is \(1\)) to set up the equation. The tangent of the angle formed by the lines from the origin to the endpoints of the latus rectum can be expressed as:
\[ \tan(45^\circ) = \frac{\frac{b^2}{a}}{ae} = 1 \]
This simplifies to:
\[ \frac{b^2}{a} = ae \]

### Step 4: Rearranging the Equation
From the equation \(\frac{b^2}{a} = ae\), we can rearrange it to find:
\[ b^2 = a^2e \]

### Step 5: Substitute into the Eccentricity Formula
The eccentricity \(e\) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting \(b^2 = a^2e\) into this formula gives:
\[ e = \sqrt{1 + \frac{a^2e}{a^2}} = \sqrt{1 + e} \]

### Step 6: Square Both Sides
Squaring both sides results in:
\[ e^2 = 1 + e \]
Rearranging this gives us a quadratic equation:
\[ e^2 - e - 1 = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula \(e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -1\), and \(c = -1\):
\[ e = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]

### Step 8: Determine the Positive Eccentricity
Since eccentricity cannot be negative, we take the positive root:
\[ e = \frac{1 + \sqrt{5}}{2} \]

### Conclusion
Thus, the eccentricity of the hyperbola whose latus rectum subtends a right angle at the center is:
\[ \boxed{\frac{1 + \sqrt{5}}{2}} \]