A hyperbola has its centre at the origin, passes through the point (4, 2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is

Options

A. `(2)/(sqrt(3))`
B. `(3)/(2)`
C. `sqrt(3)`
D. 2

Answer

To solve the problem step by step, we will follow the mathematical principles related to hyperbolas.

### Step 1: Identify the standard form of the hyperbola
The standard form of a hyperbola centered at the origin with a transverse axis along the x-axis is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 2: Determine the value of \( a \)
The length of the transverse axis is given as 4. The relationship between the length of the transverse axis and \( a \) is:
\[ \text{Length of transverse axis} = 2a \]
Thus, we have:
\[ 2a = 4 \implies a = 2 \]

### Step 3: Substitute \( a \) into the hyperbola equation
Now, substituting \( a \) into the standard form of the hyperbola, we get:
\[ \frac{x^2}{2^2} - \frac{y^2}{b^2} = 1 \implies \frac{x^2}{4} - \frac{y^2}{b^2} = 1 \]

### Step 4: Use the point (4, 2) to find \( b^2 \)
Since the hyperbola passes through the point (4, 2), we can substitute \( x = 4 \) and \( y = 2 \) into the equation:
\[ \frac{4^2}{4} - \frac{2^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{16}{4} - \frac{4}{b^2} = 1 \implies 4 - \frac{4}{b^2} = 1 \]
Rearranging gives:
\[ 4 - 1 = \frac{4}{b^2} \implies 3 = \frac{4}{b^2} \]
Cross-multiplying yields:
\[ 3b^2 = 4 \implies b^2 = \frac{4}{3} \]

### Step 5: Calculate the eccentricity \( e \)
The eccentricity \( e \) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values of \( a^2 \) and \( b^2 \):
\[ e = \sqrt{1 + \frac{\frac{4}{3}}{4}} = \sqrt{1 + \frac{4}{12}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} \]

### Final Answer
Thus, the eccentricity of the hyperbola is:
\[ \frac{2}{\sqrt{3}} \]

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