Find the eccentricity and the equations of the directrices of the ellipse `7x^(2)+16y^(2)=112`.

Answer

To find the eccentricity and the equations of the directrices of the ellipse given by the equation \(7x^2 + 16y^2 = 112\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation of the ellipse:
\[ 7x^2 + 16y^2 = 112 \]
To convert this into standard form, we divide the entire equation by 112:
\[ \frac{7x^2}{112} + \frac{16y^2}{112} = 1 \]
This simplifies to:
\[ \frac{x^2}{16} + \frac{y^2}{7} = 1 \]

### 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:
\[ a^2 = 16 \quad \text{and} \quad b^2 = 7 \]
Thus, we find:
\[ a = \sqrt{16} = 4 \quad \text{and} \quad b = \sqrt{7} \]

### Step 3: Calculate \(c\)
To find the value of \(c\) (the distance from the center to the foci), we use the formula:
\[ c = \sqrt{a^2 - b^2} \]
Substituting the values we found:
\[ c = \sqrt{16 - 7} = \sqrt{9} = 3 \]

### Step 4: Calculate the eccentricity \(e\)
The eccentricity \(e\) of the ellipse is given by:
\[ e = \frac{c}{a} \]
Substituting the values of \(c\) and \(a\):
\[ e = \frac{3}{4} \]

### Step 5: Find the equations of the directrices
The equations of the directrices for an ellipse are given by:
\[ x = \pm \frac{a}{e} \]
Substituting the values of \(a\) and \(e\):
\[ x = \pm \frac{4}{\frac{3}{4}} = \pm \left(4 \cdot \frac{4}{3}\right) = \pm \frac{16}{3} \]

### Final Result
Thus, the eccentricity of the ellipse is:
\[ \text{Eccentricity } e = \frac{3}{4} \]
And the equations of the directrices are:
\[ x = \frac{16}{3} \quad \text{and} \quad x = -\frac{16}{3} \]

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