Let `(5 tan theta , 3 sec theta)` be a point on the hyperbola for all values of `theta ne (2n + 1) pi/2` , then find the eccentricity of the hyperbola is

Options

A. `5/3`
B. `sqrt(5/3)`
C. `sqrt(34/9)`
D. `9/sqrt13`

Answer

To find the eccentricity of the hyperbola given the point \((5 \tan \theta, 3 \sec \theta)\), we can follow these steps:

### Step 1: Identify the standard form of the hyperbola
The standard form of a hyperbola can be expressed as:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
For a vertical hyperbola, the equation is:
\[ \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \]

### Step 2: Substitute the given point into the hyperbola equation
Given the point \((5 \tan \theta, 3 \sec \theta)\), we can identify:
- \(x = 5 \tan \theta\)
- \(y = 3 \sec \theta\)

### Step 3: Express \(x\) and \(y\) in terms of \(a\) and \(b\)
From the point, we can compare:
- \(x = a \tan \theta\) implies \(a = 5\)
- \(y = b \sec \theta\) implies \(b = 3\)

### Step 4: Use the formula for eccentricity
The eccentricity \(e\) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{a^2}{b^2}} \]

### Step 5: Substitute the values of \(a\) and \(b\)
Now substituting \(a = 5\) and \(b = 3\) into the eccentricity formula:
\[ e = \sqrt{1 + \frac{5^2}{3^2}} = \sqrt{1 + \frac{25}{9}} = \sqrt{1 + \frac{25}{9}} = \sqrt{\frac{9 + 25}{9}} = \sqrt{\frac{34}{9}} \]

### Step 6: Simplify the expression
This simplifies to:
\[ e = \frac{\sqrt{34}}{3} \]

### Final Answer
Thus, the eccentricity of the hyperbola is:
\[ \frac{\sqrt{34}}{3} \]