Equation of the hyperbola whose axes are the axes of coordinates (focus lying on x-axis) and which passes through the point `(-3, 1)` and has eccentricity `sqrt(3)` can be

Options

A. `x^(2)-3y^(2)=6`
B. `x^(2)-2y^(2)=7`
C. `3x^(2)-y^(2)=7`
D. `2x^(2)-y^(2)=17`

Explanation

To find the equation of the hyperbola whose axes are the coordinate axes, with a focus on the x-axis, passing through the point (-3, 1) and having an eccentricity of √3, we can follow these steps:

### Step 1: Write the standard form of the hyperbola
The standard equation of a hyperbola with a horizontal transverse axis is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 2: Use the eccentricity
The eccentricity \( e \) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Given that \( e = \sqrt{3} \), we can square both sides:
\[ 3 = 1 + \frac{b^2}{a^2} \]
This simplifies to:
\[ \frac{b^2}{a^2} = 2 \]
From this, we can express \( b^2 \) in terms of \( a^2 \):
\[ b^2 = 2a^2 \tag{1} \]

### Step 3: Substitute the point into the hyperbola equation
Since the hyperbola passes through the point (-3, 1), we substitute \( x = -3 \) and \( y = 1 \) into the hyperbola equation:
\[ \frac{(-3)^2}{a^2} - \frac{1^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{9}{a^2} - \frac{1}{b^2} = 1 \]

### Step 4: Substitute \( b^2 \) from equation (1)
Now, we substitute \( b^2 = 2a^2 \) into the equation:
\[ \frac{9}{a^2} - \frac{1}{2a^2} = 1 \]
To combine the fractions, we find a common denominator:
\[ \frac{18}{2a^2} - \frac{1}{2a^2} = 1 \]
This gives:
\[ \frac{17}{2a^2} = 1 \]

### Step 5: Solve for \( a^2 \)
Cross-multiplying gives:
\[ 17 = 2a^2 \implies a^2 = \frac{17}{2} \]

### Step 6: Find \( b^2 \)
Using equation (1):
\[ b^2 = 2a^2 = 2 \times \frac{17}{2} = 17 \]

### Step 7: Write the equation of the hyperbola
Now we can substitute \( a^2 \) and \( b^2 \) back into the standard form of the hyperbola:
\[ \frac{x^2}{\frac{17}{2}} - \frac{y^2}{17} = 1 \]
Multiplying through by the least common multiple (LCM) of the denominators, which is \( 34 \):
\[ 34 \left( \frac{x^2}{\frac{17}{2}} \right) - 34 \left( \frac{y^2}{17} \right) = 34 \]
This simplifies to:
\[ 4x^2 - 2y^2 = 34 \]
Rearranging gives:
\[ 4x^2 - 2y^2 - 34 = 0 \]

### Final Answer
The equation of the hyperbola is:
\[ 4x^2 - 2y^2 = 34 \]