Answer:
[tex]\textsf{9.} \quad \textsf{Vertices}:\;\;(0, 1) \;\; \textsf{and} \;\; (10, 1)[/tex]
[tex]\textsf{10.} \quad \textsf{Minor axis}:\;\;2\sqrt{21}[/tex]
[tex]\textsf{11.} \quad \textsf{Eccentricity}:\;\;\dfrac{2}{5}[/tex]
[tex]\textsf{12.} \quad \sf See\;attachment[/tex]
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{9.2 cm}\underline{General equation of an ellipse}\\\\$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$\\\\where:\\\phantom{ww}$\bullet$ $(h,k)$ is the center\\ \phantom{ww}$\bullet$ $(h\pm a,k)$ or $(h,k\pm b)$ are the vertices\\ \phantom{ww}$\bullet$ $(h\pm c,k)$ or $(h,k\pm c)$ are the foci where $c^2=a^2-b^2$\\\end{minipage}}[/tex]
Given:
- Foci: (3, 1) and (7, 1)
- Major axis: 10
The center of an ellipse is the midpoint of the foci.
Therefore, the center (h, k) of the ellipse is (5, 1), so:
The foci are on the major axis of an ellipse.
Given that the y-values of the given foci are the same, the ellipse is horizontal.
The foci of a horizontal ellipse are (h±c, k)
[tex]\implies 5-c=3 \implies c=2[/tex]
[tex]\implies 5+c=7 \implies c=2[/tex]
The major axis of a horizontal ellipse is 2a.
If the major axis is 10, then a = 5.
The vertices of a horizontal ellipse are (h±a, k).
Therefore the coordinates of the vertices are:
- (5-5, 1) = (0, 1)
- (5+5, 1) = (10, 1)
The minor axis of a horizontal ellipse is 2b.
As c² = a² - b² then b = √(a² - c²).
[tex]\implies b=\sqrt{5^2-2^2}=\sqrt{21}[/tex]
Therefore, the minor axis is 2√(21).
The eccentricity of the ellipse is:
[tex]\implies e=\dfrac{c}{a}=\dfrac{2}{5}[/tex]
