For questions 19-27, use the ellipse defined by the following equation: 2(x+4)^2+3(y-1)^2=24.
19. write the equation of the ellipse in standard form.
20. What are the coordinates of the center?
21. What is the length of the major axis?
22. What is the length of the minor axis?
23. What is the sum of the focal radii?
24.What are the coordinates of the foci?
25. What are the coordinates of the vertices?
26. Find the eccentricity of the ellipse.
27. Draw a sketch of the ellipse.
PLEASE SHOW WORK!!

[tex]\begin{aligned}\textsf{25.} \quad &\textsf{Vertices}:\;\;(-4-2\sqrt{3}, 1) \;\; \textsf{and}\;\; (-4+2\sqrt{3}, 1)\\ &\textsf{Co-vertices}:\;\;(-4, 1-2\sqrt{2}) \;\; \textsf{and}\;\; (-4, 1+2\sqrt{2})\end{aligned}[/tex]

[tex]\textsf{26.} \quad \textsf{Eccentricity}\;\;\dfrac{\sqrt{3}}{3}[/tex]

27. See attachment.

Step-by-step explanation:

[tex]\boxed{\begin{minipage}{5 cm}\underline{General equation of an ellipse}\\\\$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$\\\end{minipage}}[/tex]

Given equation:

[tex]2(x+4)^2+3(y-1)^2=24[/tex]

To write the equation of the ellipse in standard form, divide both sides by 24 so that the right side of the equation is equal to one:

[tex]\implies \dfrac{2(x+4)^2}{24}+\dfrac{3(y-1)^2}{24}=\dfrac{24}{24}[/tex]

[tex]\implies \dfrac{(x+4)^2}{12}+\dfrac{(y-1)^2}{8}=1[/tex]

Therefore:

[tex]h = -4[/tex]
[tex]k = 1[/tex]
[tex]a^2 = 12 \implies a=2\sqrt{3}[/tex]
[tex]b^2 = 8 \implies b=2 \sqrt{2}[/tex]

The center of the ellipse is (h, k).

Therefore, the coordinates of the center are (-4, 1).

As a > b, the ellipse is horizontal, so 2a is the major axis and 2b is the minor axis. Therefore:

Major axis = 2 × 2√3 = 4√3
Minor axis = 2 × 2√2 = 4√2

The focal radius is the distance from a point on the ellipse to a focus.

The sum of the focal radii is equal to the length of the major axis.

Therefore, the sum of the focal radii is 4√3.

As a > b, the coordinates of the foci are (h±c, k).

As c² = a² - b², first find the value of c:

[tex]\implies c^2=12-8[/tex]

[tex]\implies c=\sqrt{4}=2[/tex]

Therefore, the coordinates of the foci are:

(-4+2, 1) = (-2, 1)
(-4-2, 1) = (-6, 1)

As a > b, the coordinates of the vertices are (h±a, k):

(-4-2√3, 1) and (-4+2√3, 1)

and the coordinates of the co-vertices are (h, k±b):

(-4, 1-2√2) and (-4, 1+2√2)

The eccentricity of the ellipse is:

[tex]\implies e=\dfrac{c}{a}=\dfrac{2}{2\sqrt{3}}=\dfrac{\sqrt{3}}{3}[/tex]