What are the equations of the directrices of the ellipse `25x^(2)+16y^(2)=400`?

Options

A. `3x pm 25=0`
B. `3y pm 25=0`
C. `x pm 15 = 0`
D. `y pm 25=0`

Answer

To find the equations of the directrices of the ellipse given by the equation \( 25x^2 + 16y^2 = 400 \), we can follow these steps:

### Step 1: Write the equation in standard form
We start with the given equation of the ellipse:
\[ 25x^2 + 16y^2 = 400 \]
To convert this into standard form, we divide the entire equation by 400:
\[ \frac{25x^2}{400} + \frac{16y^2}{400} = 1 \]
This simplifies to:
\[ \frac{x^2}{16} + \frac{y^2}{25} = 1 \]

### Step 2: Identify \(a\) and \(b\)
From the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can identify:
\[ a^2 = 16 \quad \Rightarrow \quad a = 4 \]
\[ b^2 = 25 \quad \Rightarrow \quad b = 5 \]

### Step 3: Calculate the eccentricity \(e\)
The eccentricity \(e\) of the ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{a^2}{b^2}} \]
Substituting the values of \(a^2\) and \(b^2\):
\[ e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{25 - 16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Step 4: Find the equations of the directrices
The equations of the directrices for an ellipse are given by:
\[ y = \pm \frac{b}{e} \]
Substituting the values of \(b\) and \(e\):
\[ y = \pm \frac{5}{\frac{3}{5}} = \pm \frac{5 \cdot 5}{3} = \pm \frac{25}{3} \]

### Step 5: Write the final equations
Thus, the equations of the directrices are:
\[ y = \frac{25}{3} \quad \text{and} \quad y = -\frac{25}{3} \]

### Summary of the solution
The equations of the directrices of the ellipse \(25x^2 + 16y^2 = 400\) are:
\[ 3y - 25 = 0 \quad \text{and} \quad 3y + 25 = 0 \]

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