Show that the line are tangent to the given hyperbolas and determine the points of contact.
`x - 2y + 1 =0, x ^(2) -6y ^(2) =3`

Answer

To show that the line \( x - 2y + 1 = 0 \) is tangent to the hyperbola \( x^2 - 6y^2 = 3 \) and to determine the points of contact, we can follow these steps:

### Step 1: Rewrite the line equation
We start with the line equation:
\[ x - 2y + 1 = 0 \]
Rearranging this gives:
\[ x = 2y - 1 \]

**Hint:** Rearranging the line equation helps us express \( x \) in terms of \( y \), which will be useful for substitution.

### Step 2: Substitute \( x \) into the hyperbola equation
Next, we substitute \( x = 2y - 1 \) into the hyperbola equation \( x^2 - 6y^2 = 3 \):
\[ (2y - 1)^2 - 6y^2 = 3 \]

**Hint:** Substituting the expression for \( x \) into the hyperbola allows us to find the corresponding \( y \) values.

### Step 3: Expand and simplify the equation
Now, we expand the equation:
\[ (2y - 1)^2 = 4y^2 - 4y + 1 \]
Substituting this back into the hyperbola equation gives:
\[ 4y^2 - 4y + 1 - 6y^2 = 3 \]
Combining like terms results in:
\[ -2y^2 - 4y + 1 - 3 = 0 \]
This simplifies to:
\[ -2y^2 - 4y - 2 = 0 \]
Dividing the entire equation by -2 yields:
\[ y^2 + 2y + 1 = 0 \]

**Hint:** Simplifying the equation makes it easier to analyze the roots.

### Step 4: Factor the quadratic equation
The quadratic equation \( y^2 + 2y + 1 = 0 \) can be factored as:
\[ (y + 1)^2 = 0 \]
This indicates that there is a double root.

**Hint:** Recognizing that the quadratic has a double root suggests that the line is tangent to the hyperbola.

### Step 5: Solve for \( y \)
Setting \( (y + 1)^2 = 0 \) gives:
\[ y + 1 = 0 \implies y = -1 \]

**Hint:** Finding the \( y \)-coordinate of the point of contact is crucial for determining the point itself.

### Step 6: Substitute \( y \) back to find \( x \)
Now, substitute \( y = -1 \) back into the line equation to find \( x \):
\[ x = 2(-1) - 1 = -2 - 1 = -3 \]

**Hint:** Substituting back into the original line equation allows us to find the corresponding \( x \)-coordinate.

### Step 7: Determine the point of contact
Thus, the point of contact where the line is tangent to the hyperbola is:
\[ (-3, -1) \]

**Final Answer:** The line \( x - 2y + 1 = 0 \) is tangent to the hyperbola \( x^2 - 6y^2 = 3 \) at the point \( (-3, -1) \).