Find the equation of the tangents to the hyperbola `3x^(2)-4y^(2)=12` which are (i) Parallel and (ii) perpendicular to the line `y=x-7`

Answer

To find the equations of the tangents to the hyperbola \(3x^2 - 4y^2 = 12\) that are (i) parallel and (ii) perpendicular to the line \(y = x - 7\), we will follow these steps:

### Step 1: Rewrite the Hyperbola Equation
First, we rewrite the hyperbola equation in standard form.

\[ 3x^2 - 4y^2 = 12 \]

Dividing the entire equation by 12:

\[ \frac{x^2}{4} - \frac{y^2}{3} = 1 \]

This shows that \(a^2 = 4\) and \(b^2 = 3\).

### Step 2: Find the Slope of the Given Line
The equation of the line is given as \(y = x - 7\).

From this equation, we can identify the slope \(m\):

\[ m = 1 \]

### Step 3: Find Tangents Parallel to the Line
For tangents that are parallel to the line, the slope of the tangent will also be \(m = 1\).

Using the formula for the equation of the tangent to the hyperbola:

\[ y = mx + c \]

Substituting \(m = 1\):

\[ y = x + c \]

### Step 4: Use the Property of Tangency
We use the property of tangency for hyperbolas, which states:

\[ c^2 = a^2 m^2 - b^2 \]

Substituting \(a^2 = 4\), \(b^2 = 3\), and \(m = 1\):

\[ c^2 = 4(1^2) - 3 = 4 - 3 = 1 \]

Thus, \(c = \pm 1\).

### Step 5: Write the Equations of the Tangents
Substituting the values of \(c\) back into the tangent equation:

1. For \(c = 1\):
\[ y = x + 1 \]

2. For \(c = -1\):
\[ y = x - 1 \]

So, the equations of the tangents that are parallel to the line \(y = x - 7\) are:

\[ y = x + 1 \quad \text{and} \quad y = x - 1 \]

### Step 6: Find Tangents Perpendicular to the Line
For tangents that are perpendicular to the line, the product of the slopes must equal \(-1\).

Since the slope of the line is \(1\), the slope of the tangent \(m_2\) will be:

\[ m_2 = -1 \]

### Step 7: Write the Tangent Equation
Using the tangent equation again:

\[ y = mx + c \]

Substituting \(m = -1\):

\[ y = -x + c \]

### Step 8: Use the Property of Tangency Again
Using the property of tangency:

\[ c^2 = a^2 m^2 - b^2 \]

Substituting \(a^2 = 4\), \(b^2 = 3\), and \(m = -1\):

\[ c^2 = 4(-1)^2 - 3 = 4 - 3 = 1 \]

Thus, \(c = \pm 1\).

### Step 9: Write the Equations of the Tangents
Substituting the values of \(c\) back into the tangent equation:

1. For \(c = 1\):
\[ y = -x + 1 \]

2. For \(c = -1\):
\[ y = -x - 1 \]

So, the equations of the tangents that are perpendicular to the line \(y = x - 7\) are:

\[ y = -x + 1 \quad \text{and} \quad y = -x - 1 \]

### Final Answer
The equations of the tangents are:
- **Parallel to the line**: \(y = x + 1\) and \(y = x - 1\)
- **Perpendicular to the line**: \(y = -x + 1\) and \(y = -x - 1\)