The equations of the directrices of the ellipse `9x^(2) + 25y^(2) = 225` are

Options

A. `4x=pm5`
B. `16x=pm25`
C. `4x=pm25`
D. `3x=pm25`

Answer

To find the equations of the directrices of the ellipse given by the equation \(9x^2 + 25y^2 = 225\), we can follow these steps:

### Step 1: Rewrite the equation in standard form
We start by dividing the entire equation by 225 to express it in standard form.

\[ \frac{9x^2}{225} + \frac{25y^2}{225} = 1 \]

This simplifies to:

\[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]

### Step 2: Identify \(a\) and \(b\)
From the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can identify \(a^2 = 25\) and \(b^2 = 9\).

Calculating \(a\) and \(b\):

\[ a = \sqrt{25} = 5 \]
\[ b = \sqrt{9} = 3 \]

### Step 3: Calculate the eccentricity \(e\)
The eccentricity \(e\) of the ellipse is given by the formula:

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

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

\[ e = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{25 - 9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

### Step 4: Find the equations of the directrices
For an ellipse where \(a > b\), the equations of the directrices are given by:

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

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

\[ x = \pm \frac{5}{\frac{4}{5}} = \pm \frac{5 \cdot 5}{4} = \pm \frac{25}{4} \]

### Step 5: Write the final equations
Thus, the equations of the directrices are:

\[ x = \frac{25}{4} \quad \text{and} \quad x = -\frac{25}{4} \]

In a more standard form, we can express this as:

\[ 4x = \pm 25 \]

### Final Answer
The equations of the directrices of the ellipse \(9x^2 + 25y^2 = 225\) are:

\[ 4x = \pm 25 \]

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