1. Locate the vertices of an ellipse of eccentricity 0.8 whose foci lie at the points $$ (0, \pm 7) $$.

Question

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To locate the vertices of the ellipse, we need to follow these steps:

1. **Identify the foci and eccentricity**: The foci of the ellipse are given as $$ (0, \pm 7) $$, which means the foci are located at $$ (0, 7) $$ and $$ (0, -7) $$. The eccentricity $$ e $$ is given as $$ 0.8 $$.

2. **Determine the distance to the foci**: The distance from the center of the ellipse to each focus is denoted as $$ c $$. Since the foci are at $$ (0, \pm 7) $$, we have:
   $$
   c = 7
   $$

3. **Use the relationship between eccentricity, semi-major axis $$ a $$, and semi-minor axis $$ b $$**: The eccentricity $$ e $$ is defined as:
   $$
   e = \frac{c}{a}
   $$
   Rearranging this gives:
   $$
   a = \frac{c}{e}
   $$

4. **Calculate the semi-major axis $$ a $$**:
   $$
   a = \frac{7}{0.8}
   $$

5. **Calculate the semi-minor axis $$ b $$** using the relationship:
   $$
   c^2 = a^2 - b^2
   $$

Now, let's perform the calculations for $$ a $$ and $$ b $$.

First, we will calculate $$ a $$:

$$
a = \frac{7}{0.8}
$$

Next, we will calculate $$ b $$ using the equation $$ c^2 = a^2 - b^2 $$.

Let's compute these values.
Calculate the value by following steps:
- step0: Calculate:
  $$7^{2}$$
- step1: Evaluate the power:
  $$49$$
Calculate or simplify the expression $$ 7/0.8 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{7}{0.8}$$
- step1: Convert the expressions:
  $$\frac{7}{\frac{4}{5}}$$
- step2: Multiply by the reciprocal:
  $$7\times \frac{5}{4}$$
- step3: Multiply:
  $$\frac{7\times 5}{4}$$
- step4: Multiply:
  $$\frac{35}{4}$$
Calculate or simplify the expression $$ (7/0.8)^2 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\left(\frac{7}{0.8}\right)^{2}$$
- step1: Divide the numbers:
  $$\left(\frac{35}{4}\right)^{2}$$
- step2: Use the properties of exponents:
  $$\frac{35^{2}}{4^{2}}$$
- step3: Evaluate the power:
  $$\frac{1225}{4^{2}}$$
- step4: Evaluate the power:
  $$\frac{1225}{16}$$
Now we have the following calculations:

1. The semi-major axis $$ a $$:
   $$
   a = 8.75
   $$

2. The square of the distance to the foci $$ c^2 $$:
   $$
   c^2 = 49
   $$

3. The square of the semi-major axis $$ a^2 $$:
   $$
   a^2 = (8.75)^2 = \frac{1225}{16}
   $$

Now we can find $$ b^2 $$ using the relationship $$ c^2 = a^2 - b^2 $$:
$$
b^2 = a^2 - c^2
$$

Let's calculate $$ b^2 $$ and then find $$ b $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{1225}{16}-49$$
- step1: Reduce fractions to a common denominator:
  $$\frac{1225}{16}-\frac{49\times 16}{16}$$
- step2: Transform the expression:
  $$\frac{1225-49\times 16}{16}$$
- step3: Multiply the numbers:
  $$\frac{1225-784}{16}$$
- step4: Subtract the numbers:
  $$\frac{441}{16}$$
Calculate or simplify the expression $$ \sqrt(441/16) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\sqrt{\frac{441}{16}}$$
- step1: Use the properties of radicals:
  $$\frac{\sqrt{441}}{\sqrt{16}}$$
- step2: Simplify the expression:
  $$\frac{21}{4}$$
Now we have calculated $$ b^2 $$ and $$ b $$:

1. The value of $$ b^2 $$:
   $$
   b^2 = \frac{441}{16}
   $$

2. The semi-minor axis $$ b $$:
   $$
   b = \frac{21}{4} = 5.25
   $$

### Vertices of the Ellipse
Since the foci are located at $$ (0, \pm 7) $$, the ellipse is vertically oriented. The vertices are located at a distance of $$ a $$ from the center along the y-axis.

Thus, the vertices are:
$$
(0, a) \quad \text{and} \quad (0, -a)
$$
Substituting the value of $$ a $$:
$$
(0, 8.75) \quad \text{and} \quad (0, -8.75)
$$

### Final Answer
The vertices of the ellipse are located at:
$$
(0, 8.75) \quad \text{and} \quad (0, -8.75)
$$
The vertices of the ellipse are at $$ (0, 8.75) $$ and $$ (0, -8.75) $$.

Locate the vertices of an ellipse of eccentricity 0.8 whose foci lie at the points .

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