Find the equation of the tangent to the hyperbola `4x^(2)-9y^(2)=36" at "theta=pi/4`

Answer

To find the equation of the tangent to the hyperbola \(4x^2 - 9y^2 = 36\) at \(\theta = \frac{\pi}{4}\), we can follow these steps:

### Step 1: Convert the hyperbola to standard form
The given hyperbola is \(4x^2 - 9y^2 = 36\). To convert it to standard form, we divide the entire equation by 36:

\[ \frac{4x^2}{36} - \frac{9y^2}{36} = 1 \]

This simplifies to:

\[ \frac{x^2}{9} - \frac{y^2}{4} = 1 \]

From this, we can identify \(a^2 = 9\) and \(b^2 = 4\), which gives us \(a = 3\) and \(b = 2\).

### Step 2: Find the slope of the tangent line
The angle \(\theta\) is given as \(\frac{\pi}{4}\). The slope \(m\) of the tangent line is given by:

\[ m = \tan(\theta) = \tan\left(\frac{\pi}{4}\right) = 1 \]

### Step 3: Use the tangent line equation
The equation of the tangent line to the hyperbola with slope \(m\) is given by:

\[ y = mx \pm \sqrt{a^2 m^2 - b^2} \]

Substituting \(m = 1\), \(a^2 = 9\), and \(b^2 = 4\):

\[ y = 1 \cdot x \pm \sqrt{9 \cdot 1^2 - 4} \]

### Step 4: Simplify the expression
Calculating the square root term:

\[ \sqrt{9 \cdot 1 - 4} = \sqrt{9 - 4} = \sqrt{5} \]

Thus, the equation of the tangent becomes:

\[ y = x \pm \sqrt{5} \]

### Final Result
The equations of the tangents to the hyperbola at \(\theta = \frac{\pi}{4}\) are:

\[ y = x + \sqrt{5} \quad \text{and} \quad y = x - \sqrt{5} \]

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