Aliens on the planet of zorberg are creating a room that is elliptical in shape. the length of the major

10 answers
Question:

Aliens on the planet of zorberg are creating a room that is elliptical in shape. the length of the major axis is feet, and the length of the minor axis is feet. assume the center point is and that the major length is horizontal. determine the length from the center of the room to the focal points.

Answers

A. The sum of the focal radii,  is correct, just did it APEX

Step-by-step explanation:

226335 miles and 251401 miles.

c is the correct choice

because it's true

C

Step-by-step explanation:

The distance from the focus to the co-vertex is the length of the semi-major axis. Expressed in terms of the Pythagorean theorem, that relationship matches the description in selection C:

The square of the semi-minor axis is equal the the difference between the square of the semi-major axis and the square of the distance from the center to a focus.
[tex]Which of the following statements is a fact about​ ellipses? choose the correct statement below. a.[/tex]

D, 4.62

Step-by-step explanation:

e = √(1 - b²/a²)

To solve for a, first square both sides:

e² = 1 - b²/a²

Then subtract 1 from both sides:

e² - 1 = -b²/a²

Multiply both sides by -1:

1 - e² = b²/a²

Multiply by a²:

(1 - e²) a² = b²

Divide by 1 - e²:

a² = b² / (1 - e²)

Take the square root:

a = b / √(1 - e²)

So the answer is D.

When b = 4 and e = ½:

a = 4 / √(1 - (½)²)

a = 4 / √(1 - ¼)

a = 4 / √¾

a = 8 / √3

a ≈ 4.62

In the given graph, we have an horizontal ellipse . And since its horizontal, so the major axis is along x axis .

And the endpoints of major axis =(-10,3),(8,3) and of minor axis = (-1,1),(-1,5)

So the length of major axis = 8-(-10) = 18

And the length of minor axis = 5-1 = 4

It's A.

Step-by-step explanation:

The length of the major axis is a + b where a and b are the distances from each focus to any point on the ellipse.

Part 1) [tex]a=\frac{b}{\sqrt{1-e^2}}[/tex]

Part 2) [tex]a=4.62\ in[/tex]

Step-by-step explanation:

The correct question in the attached figure

Part 1)

we have the formula

[tex]e=\sqrt{1-\frac{b^2}{a^2}}[/tex]

where

e the eccentricity

a is the length of the semi-major axis

b is the length of the major axis

solve for a

That means

Isolate the variable a

[tex]e=\sqrt{1-\frac{b^2}{a^2}}[/tex]

squared both sides

[tex]e^2=1-\frac{b^2}{a^2}[/tex]

subtract 1 both sides

[tex]e^2-1=-\frac{b^2}{a^2}[/tex]

Multiply both sides by a^2

[tex](e^2-1)a^2=-b^2[/tex]

Divide both sides by (e^2-1)

[tex]a^2=\frac{-b^2}{e^2-1}[/tex]

Rewrite

[tex]a^2=\frac{-b^2}{-(1-e^2)}[/tex]

[tex]a^2=\frac{b^2}{(1-e^2)}[/tex]

square root both sides

[tex]a=\sqrt{\frac{b^2}{(1-e^2)}}[/tex]

simplify

[tex]a=\frac{b}{\sqrt{1-e^2}}[/tex]

Part 2) we have

[tex]b=4\ in\\e=1/2[/tex]

Solve for a

substitute in the formula of part 1)

[tex]a=\frac{4}{\sqrt{1-(1/2)^2}}=4.62\ in[/tex]


[tex]The formula for eccentricity, e, of an orbit is given below, where a is the length of the semi-major[/tex]

so the sun is one of the foci, and we have an eccentricity of 0.97. So the ellipse will look more or less like the picture below.


[tex]\bf \textit{ellipse, horizontal major axis}\\\\\cfrac{(x- h)^2}{ a^2}+\cfrac{(y- k)^2}{ b^2}=1\qquad \begin{cases}center\ ( h, k)\\vertices\ ( h\pm a, k)\\c=\textit{distance from}\\\qquad \textit{center to foci}\\\qquad \sqrt{ a ^2- b ^2}\\eccentricity\quad e=\cfrac{c}{a}\end{cases}\\\\-------------------------------\\\\\begin{cases}h=0\\k=0\\a=35.88\\e=0.97\end{cases}\implies \cfrac{(x-0)^2}{35.88^2}+\cfrac{(y-0)^2}{b^2}=1\\\\-------------------------------[/tex]


[tex]\bf e=\cfrac{c}{a}\implies 0.97=\cfrac{c}{35.88}\implies \boxed{34.8036=c} \\\\\\ c=\sqrt{a^2-b^2}\implies b=\sqrt{a^2-c^2}\implies b=\sqrt{35.88^2-34.8036^2} \\\\\\ \boxed{b\approx 8.72}\\\\ -------------------------------\\\\ \cfrac{(x-0)^2}{35.88^2}+\cfrac{(y-0)^2}{8.72^2}=1\implies \cfrac{x^2}{1287.3744}+\cfrac{y^2}{76.0384}=1[/tex]


[tex]Halley’s comet has an elliptical orbit with the sun at one focus. the eccentricity of the orbit is a[/tex]

Step-by-step explanation:

Given

[tex]y^2=1-3x^2[/tex]

[tex]3x^2+y^2=1[/tex]

[tex]\frac{x^2}{(\frac{1}{\sqrt{3}})^2}+\frac{y^2}{1}=1[/tex]

therefore it is a vertical ellipse

thus a=1

[tex]b=\frac{1}{\sqrt{3}}[/tex]

eccentricity of Ellipse

[tex]e^2=1-\frac{b^2}{a^2}[/tex]

[tex]e^2=1-\frac{1}{(\sqrt{3})^2}[/tex]

[tex]e^2=1-\frac{1}{3}[/tex]

[tex]e^2=\frac{2}{3}[/tex]

[tex]e=\sqrt{\frac{2}{3}}[/tex]

Focii are [tex](0,ae)[/tex] and [tex](0,-ae)[/tex]

[tex]ae=1\times \sqrt{\frac{2}{3}}[/tex]

thus focii are [tex](0,\sqrt{\frac{2}{3}})[/tex] & [tex](0,-\sqrt{\frac{2}{3}})[/tex]

(b) Length of major axis [tex]=2a=2\times 1[/tex]

length of minor axis[tex]=2b=2\times \sqrt{\frac{2}{3}}=2\cdot \sqrt{\frac{2}{3}}[/tex]

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