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

###### Question:

## 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]