Find the equation of the tagent to the hyperbola `x^(2)-4y^(2)=36` which is perpendicular to the line `x-y+4=0`.

Answer

To find the equation of the tangent to the hyperbola \( x^2 - 4y^2 = 36 \) that is perpendicular to the line \( x - y + 4 = 0 \), we can follow these steps:

### Step 1: Rewrite the hyperbola in standard form
The given equation of the hyperbola is:
\[ x^2 - 4y^2 = 36 \]
Dividing both sides by 36 gives:
\[ \frac{x^2}{36} - \frac{4y^2}{36} = 1 \]
This simplifies to:
\[ \frac{x^2}{36} - \frac{y^2}{9} = 1 \]
From this, we can identify \( a^2 = 36 \) and \( b^2 = 9 \).

### Step 2: Find the slope of the given line
The equation of the line is:
\[ x - y + 4 = 0 \]
Rearranging this gives:
\[ y = x + 4 \]
The slope \( m \) of this line is 1.

### Step 3: Determine the slope of the tangent line
Since we need the tangent line to be perpendicular to the given line, the slope of the tangent line \( m_t \) will be the negative reciprocal of the slope of the line:
\[ m_t = -1 \]

### Step 4: Use the condition for tangency
The condition for tangency for a line \( y = mx + c \) to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is given by:
\[ c^2 = a^2 m^2 - b^2 \]
Substituting \( a^2 = 36 \), \( b^2 = 9 \), and \( m = -1 \):
\[ c^2 = 36 \cdot (-1)^2 - 9 = 36 - 9 = 27 \]
Thus,
\[ c = \pm \sqrt{27} = \pm 3\sqrt{3} \]

### Step 5: Write the equation of the tangent line
The equation of the tangent line can be expressed as:
\[ y = -x + c \]
Substituting the values of \( c \):
1. For \( c = 3\sqrt{3} \):
\[ y = -x + 3\sqrt{3} \implies x + y - 3\sqrt{3} = 0 \]
2. For \( c = -3\sqrt{3} \):
\[ y = -x - 3\sqrt{3} \implies x + y + 3\sqrt{3} = 0 \]

### Final Result
Thus, the equations of the tangents to the hyperbola that are perpendicular to the given line are:
\[ x + y - 3\sqrt{3} = 0 \quad \text{and} \quad x + y + 3\sqrt{3} = 0 \]