Equation of normal to hyperbola `x^(2)-y^(2) = 2`
Passing through (6, 0)

Options

A. `(x)/(3) pm (y)/(sqrt(7)) = 2`
B. `(x)/(3) pm (y)/(sqrt(5)) = 2`
C. `(x)/(3) pm (y)/(7) = 2`
D. `(x)/(3) pm (y)/(5) = 2`

Explanation

To find the equation of the normal to the hyperbola \(x^2 - y^2 = 2\) that passes through the point (6, 0), we can follow these steps:

### Step 1: Rewrite the hyperbola in standard form
The given equation of the hyperbola is:
\[ x^2 - y^2 = 2 \]
To express it in standard form, we divide by 2:
\[ \frac{x^2}{2} - \frac{y^2}{2} = 1 \]
From this, we identify \(a^2 = 2\) and \(b^2 = 2\), which gives us \(a = \sqrt{2}\) and \(b = \sqrt{2}\).

### Step 2: Write the equation of the normal
The equation of the normal to the hyperbola at the point \((x_1, y_1)\) is given by:
\[ \frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2 \]
Substituting \(a^2\) and \(b^2\):
\[ \frac{2x}{x_1} + \frac{2y}{y_1} = 4 \]
This simplifies to:
\[ \frac{x}{x_1} + \frac{y}{y_1} = 2 \]

### Step 3: Substitute the point (6, 0)
Since the normal passes through the point (6, 0), we substitute \(x = 6\) and \(y = 0\) into the normal equation:
\[ \frac{6}{x_1} + \frac{0}{y_1} = 2 \]
This simplifies to:
\[ \frac{6}{x_1} = 2 \]
From this, we can solve for \(x_1\):
\[ x_1 = \frac{6}{2} = 3 \]

### Step 4: Substitute \(x_1\) into the hyperbola equation
Now that we have \(x_1 = 3\), we need to find \(y_1\) such that the point \((3, y_1)\) lies on the hyperbola. We substitute \(x_1\) into the hyperbola equation:
\[ x_1^2 - y_1^2 = 2 \]
Substituting \(x_1 = 3\):
\[ 3^2 - y_1^2 = 2 \]
This simplifies to:
\[ 9 - y_1^2 = 2 \]
Rearranging gives:
\[ y_1^2 = 9 - 2 = 7 \]
Thus, we find:
\[ y_1 = \pm \sqrt{7} \]

### Step 5: Write the final equation of the normal
Now we can substitute \(x_1\) and \(y_1\) back into the normal equation:
\[ \frac{x}{3} + \frac{y}{\sqrt{7}} = 2 \]
This can be rearranged to:
\[ \frac{x}{3} + \frac{y}{\sqrt{7}} = 2 \]
or
\[ x + 3\frac{y}{\sqrt{7}} = 6 \]
Thus, the equation of the normal to the hyperbola that passes through the point (6, 0) is:
\[ \frac{x}{3} + \frac{y}{\sqrt{7}} = 2 \]