Equation of directrices of `4x^(2)-9y^(2)=36` are

Options

A. `sqrt(13) x= pm3`
B. `sqrt(13 x)=pm 9`
C. `sqrt(13) x=pm 2`
D. `sqrt(13) x=pm 4`

Answer

To find the equations of the directrices of the hyperbola given by the equation \(4x^2 - 9y^2 = 36\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start by dividing the entire equation by 36 to express it in the standard form of a hyperbola.

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

This simplifies to:

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

### Step 2: Identify \(a^2\) and \(b^2\)
From the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we can identify:

\[ a^2 = 9 \quad \text{and} \quad b^2 = 4 \]

### Step 3: Calculate \(a\) and \(b\)
Now, we take the square roots to find \(a\) and \(b\):

\[ a = \sqrt{9} = 3 \quad \text{and} \quad b = \sqrt{4} = 2 \]

### Step 4: Calculate the eccentricity \(e\)
The eccentricity \(e\) of a hyperbola is given by the formula:

\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]

Substituting the values of \(b^2\) and \(a^2\):

\[ e = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{9 + 4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} \]

### Step 5: Find the equations of the directrices
The equations of the directrices for a hyperbola are given by:

\[ x = \pm \frac{a}{e} \]

Substituting the values of \(a\) and \(e\):

\[ x = \pm \frac{3}{\frac{\sqrt{13}}{3}} = \pm \frac{3 \cdot 3}{\sqrt{13}} = \pm \frac{9}{\sqrt{13}} \]

### Step 6: Final form of the directrices
Thus, the equations of the directrices are:

\[ x = \frac{9}{\sqrt{13}} \quad \text{and} \quad x = -\frac{9}{\sqrt{13}} \]

### Summary of the Solution
The equations of the directrices of the hyperbola \(4x^2 - 9y^2 = 36\) are:

\[ x = \frac{9}{\sqrt{13}} \quad \text{and} \quad x = -\frac{9}{\sqrt{13}} \]