To find the equations of the directrices for the given ellipses, we will follow these steps for each ellipse:
1. Identify \( a^2 \) and \( b^2 \) from the given equation of the ellipse.
2. Determine whether \( a^2 > b^2 \) or \( b^2 > a^2 \) to decide which case to apply.
3. Calculate the eccentricity \( e \) using the appropriate formula.
4. Write the equation of the directrices based on the eccentricity and the values of \( a \) and \( b \).
Let's solve each ellipse step by step.
### (i) For the ellipse \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \)
1. Identify \( a^2 = 25 \) and \( b^2 = 9 \). Here, \( a = 5 \) and \( b = 3 \).
2. Since \( a^2 > b^2 \), we use the formula for the directrix: \( x = \pm \frac{a}{e} \).
3. Calculate eccentricity \( e \):
\[
e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}
\]
4. Now, substitute \( a \) and \( e \) into the directrix equation:
\[
x = \pm \frac{5}{\frac{4}{5}} = \pm \frac{5 \cdot 5}{4} = \pm \frac{25}{4}
\]
So, the equations of the directrices are:
\[
x = \frac{25}{4} \quad \text{and} \quad x = -\frac{25}{4}
\]
### (ii) For the ellipse \( \frac{x^2}{36} + \frac{y^2}{16} = 1 \)
1. Identify \( a^2 = 36 \) and \( b^2 = 16 \). Here, \( a = 6 \) and \( b = 4 \).
2. Since \( a^2 > b^2 \), we use the formula for the directrix: \( x = \pm \frac{a}{e} \).
3. Calculate eccentricity \( e \):
\[
e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{36}} = \sqrt{\frac{20}{36}} = \frac{\sqrt{20}}{6} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}
\]
4. Now, substitute \( a \) and \( e \) into the directrix equation:
\[
x = \pm \frac{6}{\frac{\sqrt{5}}{3}} = \pm \frac{6 \cdot 3}{\sqrt{5}} = \pm \frac{18}{\sqrt{5}}
\]
So, the equations of the directrices are:
\[
x = \frac{18}{\sqrt{5}} \quad \text{and} \quad x = -\frac{18}{\sqrt{5}}
\]
### (iii) For the ellipse \( \frac{x^2}{49} + \frac{y^2}{36} = 1 \)
1. Identify \( a^2 = 49 \) and \( b^2 = 36 \). Here, \( a = 7 \) and \( b = 6 \).
2. Since \( a^2 > b^2 \), we use the formula for the directrix: \( x = \pm \frac{a}{e} \).
3. Calculate eccentricity \( e \):
\[
e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{36}{49}} = \sqrt{\frac{13}{49}} = \frac{\sqrt{13}}{7}
\]
4. Now, substitute \( a \) and \( e \) into the directrix equation:
\[
x = \pm \frac{7}{\frac{\sqrt{13}}{7}} = \pm \frac{7 \cdot 7}{\sqrt{13}} = \pm \frac{49}{\sqrt{13}}
\]
So, the equations of the directrices are:
\[
x = \frac{49}{\sqrt{13}} \quad \text{and} \quad x = -\frac{49}{\sqrt{13}}
\]
### (iv) For the ellipse \( \frac{x^2}{100} + \frac{y^2}{400} = 1 \)
1. Identify \( a^2 = 100 \) and \( b^2 = 400 \). Here, \( a = 10 \) and \( b = 20 \).
2. Since \( b^2 > a^2 \), we use the formula for the directrix: \( y = \pm \frac{b}{e} \).
3. Calculate eccentricity \( e \):
\[
e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{100}{400}} = \sqrt{\frac{300}{400}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}
\]
4. Now, substitute \( b \) and \( e \) into the directrix equation:
\[
y = \pm \frac{20}{\frac{\sqrt{3}}{2}} = \pm \frac{20 \cdot 2}{\sqrt{3}} = \pm \frac{40}{\sqrt{3}}
\]
So, the equations of the directrices are:
\[
y = \frac{40}{\sqrt{3}} \quad \text{and} \quad y = -\frac{40}{\sqrt{3}}
\]
### Summary of Directrices
1. For \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \): \( x = \frac{25}{4}, x = -\frac{25}{4} \)
2. For \( \frac{x^2}{36} + \frac{y^2}{16} = 1 \): \( x = \frac{18}{\sqrt{5}}, x = -\frac{18}{\sqrt{5}} \)
3. For \( \frac{x^2}{49} + \frac{y^2}{36} = 1 \): \( x = \frac{49}{\sqrt{13}}, x = -\frac{49}{\sqrt{13}} \)
4. For \( \frac{x^2}{100} + \frac{y^2}{400} = 1 \): \( y = \frac{40}{\sqrt{3}}, y = -\frac{40}{\sqrt{3}} \)
