Mdthbs

Will the volt, ampere, ohm or other electrical units change on May 20th, 2019?

With today's technology, could iron be smelted at La Rinconada?

What is this weird d12 for?

How could it be that 80% of townspeople were farmers during the Edo period in Japan?

Is there any deeper thematic meaning to the white horse that Arya finds in The Bells (S08E05)?

What do the "optional" resistor and capacitor do in this circuit?

Do people who work at research institutes consider themselves "academics"?

Which creature is depicted in this Xanathar's Guide illustration of a war mage?

Why is the Advance Variation considered strong vs the Caro-Kann but not vs the Scandinavian?

Formal Definition of Dot Product

What dog breeds survive the apocalypse for generations?

How do I know which cipher suites can be disabled?

Should generated documentation be stored in a Git repository?

Network latencies between opposite ends of the Earth

​Cuban​ ​Primes

Is my test coverage up to snuff?

Slice a list based on an index and items behind it in python

How to redirect stdout to a file, and stdout+stderr to another one?

tikz drawing rectangle discretized with triangle lattices and its centroids

Was the dragon prowess intentionally downplayed in S08E04?

Does it matter what way the tires go if no directional arrow?

What was Varys trying to do at the beginning of S08E05?

Can anyone give me examples of the relative-determinative 'which'?

Do crew rest seats count towards the maximum allowed number of seats per flight attendant?

Find the equation and height of an elliptical whispering room

Find equation for hyperbolaGeneralization of ellipse equation to higher dimensional surfacesGetting the equation of an ellipse using the constant and the fociEllipse in Cartesian and in Polar CoordinatesHelp creating equation for parabola word problem?Location of the foci of a hyperbola as the value of $a$ becomes increasingly smaller than the value of $b$Having some problems with understanding conics and graphing (eccentricity)Parabola ProblemFind the Vertices of an Ellipse Given Its Foci and Distance Between VerticesEllipse: Known Distance from Focus to Far Side $(A+C)$ and $B$

3

$begingroup$

The room is 150 feet long and the distance from the center of the room to the foci is 60 feet.

Finding $a^2$ is easy its
$$2a=150$$
$$a=75$$
$$a^2=5625$$

but where I get lost is finding $b^2$, I know I shouldn't look at the answer before solving but I was stuck for 10 mins. trying to figure it out.

Anyway $b=45$ and I have no idea how to calucate that because what am I suppose to use the $60$ feet from the foci for?

I thought it was
$$2b=60$$
$$b=30$$
$$b^2=900$$
but that doesn't work, then I thought it should be
$$2b=120$$
$$b=60$$
$$b^2=3600$$
and again I get stuck.

conic-sections elliptic-curves

share|cite|improve this question

asked 2 hours ago

Eric BrownEric Brown

1087

$endgroup$

add a comment | 

3

$begingroup$

The room is 150 feet long and the distance from the center of the room to the foci is 60 feet.

Finding $a^2$ is easy its
$$2a=150$$
$$a=75$$
$$a^2=5625$$

but where I get lost is finding $b^2$, I know I shouldn't look at the answer before solving but I was stuck for 10 mins. trying to figure it out.

Anyway $b=45$ and I have no idea how to calucate that because what am I suppose to use the $60$ feet from the foci for?

I thought it was
$$2b=60$$
$$b=30$$
$$b^2=900$$
but that doesn't work, then I thought it should be
$$2b=120$$
$$b=60$$
$$b^2=3600$$
and again I get stuck.

conic-sections elliptic-curves

share|cite|improve this question

asked 2 hours ago

Eric BrownEric Brown

1087

$endgroup$

add a comment | 

$begingroup$

The room is 150 feet long and the distance from the center of the room to the foci is 60 feet.

Finding $a^2$ is easy its
$$2a=150$$
$$a=75$$
$$a^2=5625$$

but where I get lost is finding $b^2$, I know I shouldn't look at the answer before solving but I was stuck for 10 mins. trying to figure it out.

Anyway $b=45$ and I have no idea how to calucate that because what am I suppose to use the $60$ feet from the foci for?

I thought it was
$$2b=60$$
$$b=30$$
$$b^2=900$$
but that doesn't work, then I thought it should be
$$2b=120$$
$$b=60$$
$$b^2=3600$$
and again I get stuck.

conic-sections elliptic-curves

share|cite|improve this question

asked 2 hours ago

Eric BrownEric Brown

1087

$endgroup$

share|cite|improve this question

asked 2 hours ago

Eric BrownEric Brown

1087

share|cite|improve this question

asked 2 hours ago

Eric BrownEric Brown

1087

asked 2 hours ago

Eric BrownEric Brown

1087

asked 2 hours ago

Eric BrownEric Brown

1087

3 Answers
3

active

oldest

votes

4

$begingroup$

Consider this sketch

We know

• $BF_1+BF_2= AF_1+AF_2 = 2AO = 150$ as this is an ellipse with a fixed combined distance from the foci to a point on the edge




• and that $OF_1=OF_2=60$




• so $OA^2 =75^2$ as you found




• while $OB^2 = 75^2-60^2 = 45^2$ by Pythagoras

so the height (or width?) is $2OB=90$

and the equation of the room might be $dfrac{x^2}{75^2}+dfrac{y^2}{45^2}=1$

share|cite|improve this answer

answered 1 hour ago

HenryHenry

103k483172

$endgroup$

add a comment | 

1

$begingroup$

You can use $c^2=a^2-b^2$, so $b=sqrt{a^2-c^2}=sqrt{75^2-60^2}=15sqrt{5^2-4^2}=15cdot3=45$

share|cite|improve this answer

answered 1 hour ago

AndreiAndrei

14.4k21330

$endgroup$

add a comment | 

1

$begingroup$

It is simple, just you need to know more about ellipse. In ellipse, the distance of centre from focus is $ae$, where $e=sqrt{1-(frac b a)^2}$. Now it is easy.

e is called eccentricity of the ellipse.

share|cite|improve this answer

answered 1 hour ago

TojrahTojrah

1,068212

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3226210%2ffind-the-equation-and-height-of-an-elliptical-whispering-room%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

Consider this sketch

We know

• $BF_1+BF_2= AF_1+AF_2 = 2AO = 150$ as this is an ellipse with a fixed combined distance from the foci to a point on the edge




• and that $OF_1=OF_2=60$




• so $OA^2 =75^2$ as you found




• while $OB^2 = 75^2-60^2 = 45^2$ by Pythagoras

so the height (or width?) is $2OB=90$

and the equation of the room might be $dfrac{x^2}{75^2}+dfrac{y^2}{45^2}=1$

share|cite|improve this answer

answered 1 hour ago

HenryHenry

103k483172

$endgroup$

add a comment | 

4

$begingroup$

Consider this sketch

We know

• $BF_1+BF_2= AF_1+AF_2 = 2AO = 150$ as this is an ellipse with a fixed combined distance from the foci to a point on the edge




• and that $OF_1=OF_2=60$




• so $OA^2 =75^2$ as you found




• while $OB^2 = 75^2-60^2 = 45^2$ by Pythagoras

so the height (or width?) is $2OB=90$

and the equation of the room might be $dfrac{x^2}{75^2}+dfrac{y^2}{45^2}=1$

share|cite|improve this answer

answered 1 hour ago

HenryHenry

103k483172

$endgroup$

add a comment | 

$begingroup$

Consider this sketch

We know

• $BF_1+BF_2= AF_1+AF_2 = 2AO = 150$ as this is an ellipse with a fixed combined distance from the foci to a point on the edge




• and that $OF_1=OF_2=60$




• so $OA^2 =75^2$ as you found




• while $OB^2 = 75^2-60^2 = 45^2$ by Pythagoras

so the height (or width?) is $2OB=90$

and the equation of the room might be $dfrac{x^2}{75^2}+dfrac{y^2}{45^2}=1$

share|cite|improve this answer

answered 1 hour ago

HenryHenry

103k483172

$endgroup$

share|cite|improve this answer

answered 1 hour ago

HenryHenry

103k483172

answered 1 hour ago

HenryHenry

103k483172

answered 1 hour ago

HenryHenry

103k483172

1

$begingroup$

You can use $c^2=a^2-b^2$, so $b=sqrt{a^2-c^2}=sqrt{75^2-60^2}=15sqrt{5^2-4^2}=15cdot3=45$

share|cite|improve this answer

answered 1 hour ago

AndreiAndrei

14.4k21330

$endgroup$

add a comment | 

1

$begingroup$

You can use $c^2=a^2-b^2$, so $b=sqrt{a^2-c^2}=sqrt{75^2-60^2}=15sqrt{5^2-4^2}=15cdot3=45$

share|cite|improve this answer

answered 1 hour ago

AndreiAndrei

14.4k21330

$endgroup$

add a comment | 

$begingroup$

You can use $c^2=a^2-b^2$, so $b=sqrt{a^2-c^2}=sqrt{75^2-60^2}=15sqrt{5^2-4^2}=15cdot3=45$

share|cite|improve this answer

answered 1 hour ago

AndreiAndrei

14.4k21330

$endgroup$

share|cite|improve this answer

answered 1 hour ago

AndreiAndrei

14.4k21330

answered 1 hour ago

AndreiAndrei

14.4k21330

answered 1 hour ago

AndreiAndrei

14.4k21330

1

$begingroup$

It is simple, just you need to know more about ellipse. In ellipse, the distance of centre from focus is $ae$, where $e=sqrt{1-(frac b a)^2}$. Now it is easy.

e is called eccentricity of the ellipse.

share|cite|improve this answer

answered 1 hour ago

TojrahTojrah

1,068212

$endgroup$

add a comment | 

1

$begingroup$

It is simple, just you need to know more about ellipse. In ellipse, the distance of centre from focus is $ae$, where $e=sqrt{1-(frac b a)^2}$. Now it is easy.

e is called eccentricity of the ellipse.

share|cite|improve this answer

answered 1 hour ago

TojrahTojrah

1,068212

$endgroup$

add a comment | 

$begingroup$

It is simple, just you need to know more about ellipse. In ellipse, the distance of centre from focus is $ae$, where $e=sqrt{1-(frac b a)^2}$. Now it is easy.

e is called eccentricity of the ellipse.

share|cite|improve this answer

answered 1 hour ago

TojrahTojrah

1,068212

$endgroup$

share|cite|improve this answer

answered 1 hour ago

TojrahTojrah

1,068212

answered 1 hour ago

TojrahTojrah

1,068212

answered 1 hour ago

TojrahTojrah

1,068212