52. Vertices: $$ (0, \pm 8) $$; eccentricity: $$ \frac{1}{2} $$ 33. Foci: $$ (1,1),(1,13) $$; eccentricity: $$ \frac{2}{3} $$ 34. Foci: $$ (-6,5),(2,5) $$; eccentricity: $$ \frac{4}{5} $$

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**52. Vertices: $$ (0,\pm8) $$; eccentricity: $$ \frac{1}{2} $$**

1. The ellipse has center at the origin, and the vertices along the $$y$$–axis indicate that the major axis is vertical. Thus, the vertices are $$(0, \pm a)$$ with
   $$
   a = 8.
   $$
2. For an ellipse, the eccentricity is given by 
   $$
   e=\frac{c}{a}.
   $$
   With $$e=\frac{1}{2}$$, we have:
   $$
   \frac{c}{8}=\frac{1}{2}\quad\Longrightarrow\quad c=4.
   $$
3. The relationship among the semi–axis lengths is 
   $$
   b^2=a^2-c^2.
   $$
   Thus,
   $$
   b^2=8^2-4^2=64-16=48.
   $$
4. The equation of the ellipse (vertical major axis) is
   $$
   \frac{x^2}{48}+\frac{y^2}{64}=1.
   $$

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**33. Foci: $$ (1,1) $$ and $$ (1,13) $$; eccentricity: $$\frac{2}{3}$$**

1. The foci lie on the vertical line $$x=1$$. The center is the midpoint of the foci:
   $$
   \left(1,\frac{1+13}{2}\right)= (1,7).
   $$
2. The distance from the center to a focus is 
   $$
   c=|7-1|=6.
   $$
3. For an ellipse, the eccentricity is 
   $$
   e=\frac{c}{a}.
   $$
   With $$e=\frac{2}{3}$$, we solve for $$a$$:
   $$
   \frac{6}{a}=\frac{2}{3}\quad\Longrightarrow\quad a=\frac{6\cdot 3}{2}=9.
   $$
4. Compute $$b$$ using 
   $$
   b^2=a^2-c^2=9^2-6^2=81-36=45.
   $$
5. Since the major axis is vertical, the standard form of the ellipse is
   $$
   \frac{(x-1)^2}{45}+\frac{(y-7)^2}{81}=1.
   $$

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**34. Foci: $$ (-6,5) $$ and $$ (2,5) $$; eccentricity: $$\frac{4}{5}$$**

1. The foci lie on the horizontal line $$y=5$$. The center is the midpoint of the foci:
   $$
   \left(\frac{-6+2}{2},\, 5\right)=(-2,5).
   $$
2. The distance from the center to a focus is 
   $$
   c=|-2-(-6)|=4 \quad \text{(or } |2-(-2)|=4\text{)}.
   $$
3. Using the ellipse definition $$e=\frac{c}{a}$$ with $$e=\frac{4}{5}$$:
   $$
   \frac{4}{a}=\frac{4}{5}\quad\Longrightarrow\quad a=5.
   $$
4. Compute $$b$$ using 
   $$
   b^2=a^2-c^2=5^2-4^2=25-16=9.
   $$
5. Since the major axis is horizontal, the equation of the ellipse is
   $$
   \frac{(x+2)^2}{25}+\frac{(y-5)^2}{9}=1.
   $$
**52. Vertices: $$ (0,\pm8) $$; Eccentricity: $$ \frac{1}{2} $$** - Equation of the ellipse: $$ \frac{x^2}{48} + \frac{y^2}{64} = 1 $$ **33. Foci: $$ (1,1) $$ and $$ (1,13) $$; Eccentricity: $$ \frac{2}{3} $$** - Equation of the ellipse: $$ \frac{(x-1)^2}{45} + \frac{(y-7)^2}{81} = 1 $$ **34. Foci: $$ (-6,5) $$ and $$ (2,5) $$; Eccentricity: $$ \frac{4}{5} $$** - Equation of the ellipse: $$ \frac{(x+2)^2}{25} + \frac{(y-5)^2}{9} = 1 $$

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