Part 3: Write the equation of and graph an ellipse.
Given the foci and vertices of an ellipse, complete the following steps.
Vertices at (4, -6 +-9)
Foci at (4, -6 +-5sqrt2)

a) What is the center of the ellipse? Explain how to determine this from the given points.

b) Write the standard equation of the ellipse. (3 points

c) What are the lengths of the major and minor axes? (2 points)
Major Axis:
Minor Axis:

d) Sketch a graph of the ellipse and label the center, foci, and the major and minor axes. (4 points)

Answer:

Step-by-step explanation:

The center is halfway between vertices, at (4, -6).

It is also halfway between foci.

:::::

The vertices are vertically aligned, so the parabola is vertical.

General equation for a vertical ellipse:

 (y-k)²/a² + (x-h)²/b² = 1

with

 center (h,k)

 vertices (h,k±a)

 co-vertices (h±b,k)

 foci (h,k±c), c² = a²-b²

Apply your data and solve for h, k, a, and b.

center (h,k) = (4, -6)

h = 4

k = -6

vertices (4,-6±a) = (4,-6±9)

a = 9

foci (4,-6±c) = (4,-6±5√2)

b² = a² - c² = 9² - (5√2)² = 31

b = √31

The equation becomes

 (y+6)²/81 + (x-4)²/31 = 1

:::::

length of major axis = 2a = 18

length of minor axis = 2b = 2√31

The various parts of the question needs to be answered.

Center: (4,-6)

Standard equation: [tex]\dfrac{(x-4)^2}{\sqrt{31}^2}+\dfrac{(y+6)^2}{9^2}=1[/tex]

Length of major axis: 18 units

Length of minor axis: [tex]2\sqrt{31}\ \text{units}[/tex]

Graph is attached.

The Vertices are of the form

[tex](h,k\pm a)=(4,-6\pm 9)[/tex]

This means that the ellipse is symmetrical to the y axis.

The center of an ellipse is given by [tex](h,k)[/tex]

So, here the center is [tex](4,-6)[/tex]

[tex]a=\pm 9[/tex]

The foci are of the form

[tex](h,k\pm c)=(4,-6\pm 5\sqrt{2})[/tex]

So, [tex]c=\pm 5\sqrt{2}[/tex]

[tex]b=\sqrt{a^2-c^2}\\\Rightarrow b=\sqrt{9^2-(5\sqrt{2})^2}\\\Rightarrow b=\pm\sqrt{31}[/tex]

The standard equation for major axis being parallel to y axis is given by

[tex]\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1\\\Rightarrow \dfrac{(x-4)^2}{\sqrt{31}^2}+\dfrac{(y+6)^2}{9^2}=1[/tex]

Length of major axis is

[tex]2a=2\times 9\\ =18\ \text{units}[/tex]

Length of minor axis is

[tex]2b=2\sqrt{31}\ \text{units}[/tex]

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