Find the equations of the tangents to the hyperbla `4x ^(2) - 9y ^(2) = 144,` which are perpendicular to the line `6x + 5y = 21.`

Answer

To find the equations of the tangents to the hyperbola \(4x^2 - 9y^2 = 144\) that are perpendicular to the line \(6x + 5y = 21\), we will follow these steps:

### Step 1: Rewrite the hyperbola in standard form
The given hyperbola is:
\[ 4x^2 - 9y^2 = 144 \]
Dividing through by 144, we get:
\[ \frac{x^2}{36} - \frac{y^2}{16} = 1 \]
This shows that \(a^2 = 36\) and \(b^2 = 16\), hence \(a = 6\) and \(b = 4\).

### Step 2: Find the slope of the given line
The equation of the line is:
\[ 6x + 5y = 21 \]
Rearranging it into slope-intercept form \(y = mx + c\):
\[ 5y = -6x + 21 \implies y = -\frac{6}{5}x + \frac{21}{5} \]
Thus, the slope \(m\) of the line is \(-\frac{6}{5}\).

### Step 3: Determine the slope of the tangents
Since the tangents are perpendicular to the line, the slope of the tangents \(m_t\) can be found using the negative reciprocal:
\[ m_t = -\frac{1}{m} = -\frac{1}{-\frac{6}{5}} = \frac{5}{6} \]

### Step 4: Write the equation of the tangent
The equation of the tangent to the hyperbola with slope \(m\) is given by:
\[ y = mx \pm \sqrt{m^2 a^2 - b^2} \]
Substituting \(m = \frac{5}{6}\), \(a = 6\), and \(b = 4\):
\[ y = \frac{5}{6}x \pm \sqrt{\left(\frac{5}{6}\right)^2 \cdot 36 - 16} \]
Calculating \(m^2 a^2\):
\[ \left(\frac{5}{6}\right)^2 \cdot 36 = \frac{25}{36} \cdot 36 = 25 \]
Now, substituting this back:
\[ y = \frac{5}{6}x \pm \sqrt{25 - 16} = \frac{5}{6}x \pm \sqrt{9} = \frac{5}{6}x \pm 3 \]

### Step 5: Write the final equations of the tangents
Thus, we have two equations for the tangents:
1. \(y = \frac{5}{6}x + 3\)
2. \(y = \frac{5}{6}x - 3\)

Rearranging these into standard form:
1. \(5x - 6y + 18 = 0\)
2. \(5x - 6y - 18 = 0\)

### Final Answer
The equations of the tangents to the hyperbola that are perpendicular to the given line are:
\[ 5x - 6y + 18 = 0 \quad \text{and} \quad 5x - 6y - 18 = 0 \]