How do I find the length of the major and minor axes of the ellipse $frac{x^2}{49}+frac{y^2}{25}=1$?












1












$begingroup$


How do I find the length of the major and minor axes of the equation $frac{x^2}{49}+frac{y^2}{25}$?
For the vertices I got $(7,0)$ and $(-7,0)$
The foci: $(2sqrt{6},0)$, $(-2sqrt{6},0)$
Eccentricity: $frac{2sqrt{6}}{7}$
I'm not sure if those are correct, but I'm also confused on what the major and minor axes are.
$a$ is $7$, and $b$ is $5$, but I don't know if those would be the right values.






algebra-precalculus conic-sections







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Just to be a pedant, you don't have an equation.
    $endgroup$
    – user608030
    Dec 13 '18 at 16:22






  • 1




    $begingroup$
    ${x^2 over 49} + {y^2 over 25}$ is not an equation.
    $endgroup$
    – David G. Stork
    Dec 13 '18 at 16:22
















1












$begingroup$


How do I find the length of the major and minor axes of the equation $frac{x^2}{49}+frac{y^2}{25}$?
For the vertices I got $(7,0)$ and $(-7,0)$
The foci: $(2sqrt{6},0)$, $(-2sqrt{6},0)$
Eccentricity: $frac{2sqrt{6}}{7}$
I'm not sure if those are correct, but I'm also confused on what the major and minor axes are.
$a$ is $7$, and $b$ is $5$, but I don't know if those would be the right values.






algebra-precalculus conic-sections







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Just to be a pedant, you don't have an equation.
    $endgroup$
    – user608030
    Dec 13 '18 at 16:22






  • 1




    $begingroup$
    ${x^2 over 49} + {y^2 over 25}$ is not an equation.
    $endgroup$
    – David G. Stork
    Dec 13 '18 at 16:22














1












1








1





$begingroup$


How do I find the length of the major and minor axes of the equation $frac{x^2}{49}+frac{y^2}{25}$?
For the vertices I got $(7,0)$ and $(-7,0)$
The foci: $(2sqrt{6},0)$, $(-2sqrt{6},0)$
Eccentricity: $frac{2sqrt{6}}{7}$
I'm not sure if those are correct, but I'm also confused on what the major and minor axes are.
$a$ is $7$, and $b$ is $5$, but I don't know if those would be the right values.






algebra-precalculus conic-sections







share|cite|improve this question











$endgroup$




How do I find the length of the major and minor axes of the equation $frac{x^2}{49}+frac{y^2}{25}$?
For the vertices I got $(7,0)$ and $(-7,0)$
The foci: $(2sqrt{6},0)$, $(-2sqrt{6},0)$
Eccentricity: $frac{2sqrt{6}}{7}$
I'm not sure if those are correct, but I'm also confused on what the major and minor axes are.
$a$ is $7$, and $b$ is $5$, but I don't know if those would be the right values.






algebra-precalculus conic-sections




algebra-precalculus conic-sections






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 17:01









Martin Sleziak

45k10122277




45k10122277










asked Dec 13 '18 at 16:04









User231User231

105




105








  • 1




    $begingroup$
    Just to be a pedant, you don't have an equation.
    $endgroup$
    – user608030
    Dec 13 '18 at 16:22






  • 1




    $begingroup$
    ${x^2 over 49} + {y^2 over 25}$ is not an equation.
    $endgroup$
    – David G. Stork
    Dec 13 '18 at 16:22














  • 1




    $begingroup$
    Just to be a pedant, you don't have an equation.
    $endgroup$
    – user608030
    Dec 13 '18 at 16:22






  • 1




    $begingroup$
    ${x^2 over 49} + {y^2 over 25}$ is not an equation.
    $endgroup$
    – David G. Stork
    Dec 13 '18 at 16:22








1




1




$begingroup$
Just to be a pedant, you don't have an equation.
$endgroup$
– user608030
Dec 13 '18 at 16:22




$begingroup$
Just to be a pedant, you don't have an equation.
$endgroup$
– user608030
Dec 13 '18 at 16:22




1




1




$begingroup$
${x^2 over 49} + {y^2 over 25}$ is not an equation.
$endgroup$
– David G. Stork
Dec 13 '18 at 16:22




$begingroup$
${x^2 over 49} + {y^2 over 25}$ is not an equation.
$endgroup$
– David G. Stork
Dec 13 '18 at 16:22










2 Answers
2






active

oldest

votes


















3












$begingroup$

Based on the context that you provide, you're likely dealing with an ellipse whose equation is $${x^2over49} + {y^2over25} color{red}{= 1}.$$ The standard-form of an ellipse whose center is at the origin is given by $${x^2over a^2} + {y^2over b^2} = 1,$$ where $a$ and $b$ are the length of the horizontal and vertical axes respectively. Depending on whether $a$ is bigger or $b$ is bigger determines which one is the major axis. In this case, $a = 7$ and $b = 5$ by simple square root. We clearly see that $a = 7$ is bigger; since that is associated with the $x$ variable, you know that you have horizontal major axis, and thus you have the following vertices of the ellipse: $$(pm a, 0), (0,pm b) quad = quad (pm 7, 0), (pm 5, 0).$$ Geometrically speaking, this just means we go out $7$ both left and right from the center, and $5$ up and down from the center. Being able to interpret the ellipse geometrically in effect makes these kinds of problems much easier, as you can help reduce the number of formulas that you need to memorize.




This is especially true when you begin to shift the center to some general $(h,k)$; then the vertices become $(h pm a, k), (h, kpm b)$. This is a lot more "specific stuff" to remember, but if you interpret it geometrically, there is less math you have to do and less formulas required to memorize.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you! Sorry I forgot to add the =1. Are my other values correct?
    $endgroup$
    – User231
    Dec 13 '18 at 16:36










  • $begingroup$
    The values appear correct to me!
    $endgroup$
    – Decaf-Math
    Dec 13 '18 at 16:39



















1












$begingroup$

There's a lot going on here. First, you're missing an $=1$ in your equations.



The major axis is the longest "diameter" you can draw in the ellipse. The minor axis is the shortest. $a=7$ is what people would call the length of the major SEMI-axis.
So the length of the major axis is $14$.



It might help to plot the $x$ and $y$ intercepts (which you get by plugging in $y=0$ and $x=0$, respectively.) Those will be the endpoint of the axes you're looking for.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    So if the major axis is 7(2)=14, then the minor axis would be 5(2)=10?
    $endgroup$
    – User231
    Dec 13 '18 at 16:44










  • $begingroup$
    Yes. See mathopenref.com/ellipseaxes.html
    $endgroup$
    – B. Goddard
    Dec 13 '18 at 17:03












Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038207%2fhow-do-i-find-the-length-of-the-major-and-minor-axes-of-the-ellipse-fracx2%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Based on the context that you provide, you're likely dealing with an ellipse whose equation is $${x^2over49} + {y^2over25} color{red}{= 1}.$$ The standard-form of an ellipse whose center is at the origin is given by $${x^2over a^2} + {y^2over b^2} = 1,$$ where $a$ and $b$ are the length of the horizontal and vertical axes respectively. Depending on whether $a$ is bigger or $b$ is bigger determines which one is the major axis. In this case, $a = 7$ and $b = 5$ by simple square root. We clearly see that $a = 7$ is bigger; since that is associated with the $x$ variable, you know that you have horizontal major axis, and thus you have the following vertices of the ellipse: $$(pm a, 0), (0,pm b) quad = quad (pm 7, 0), (pm 5, 0).$$ Geometrically speaking, this just means we go out $7$ both left and right from the center, and $5$ up and down from the center. Being able to interpret the ellipse geometrically in effect makes these kinds of problems much easier, as you can help reduce the number of formulas that you need to memorize.




This is especially true when you begin to shift the center to some general $(h,k)$; then the vertices become $(h pm a, k), (h, kpm b)$. This is a lot more "specific stuff" to remember, but if you interpret it geometrically, there is less math you have to do and less formulas required to memorize.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you! Sorry I forgot to add the =1. Are my other values correct?
    $endgroup$
    – User231
    Dec 13 '18 at 16:36










  • $begingroup$
    The values appear correct to me!
    $endgroup$
    – Decaf-Math
    Dec 13 '18 at 16:39
















3












$begingroup$

Based on the context that you provide, you're likely dealing with an ellipse whose equation is $${x^2over49} + {y^2over25} color{red}{= 1}.$$ The standard-form of an ellipse whose center is at the origin is given by $${x^2over a^2} + {y^2over b^2} = 1,$$ where $a$ and $b$ are the length of the horizontal and vertical axes respectively. Depending on whether $a$ is bigger or $b$ is bigger determines which one is the major axis. In this case, $a = 7$ and $b = 5$ by simple square root. We clearly see that $a = 7$ is bigger; since that is associated with the $x$ variable, you know that you have horizontal major axis, and thus you have the following vertices of the ellipse: $$(pm a, 0), (0,pm b) quad = quad (pm 7, 0), (pm 5, 0).$$ Geometrically speaking, this just means we go out $7$ both left and right from the center, and $5$ up and down from the center. Being able to interpret the ellipse geometrically in effect makes these kinds of problems much easier, as you can help reduce the number of formulas that you need to memorize.




This is especially true when you begin to shift the center to some general $(h,k)$; then the vertices become $(h pm a, k), (h, kpm b)$. This is a lot more "specific stuff" to remember, but if you interpret it geometrically, there is less math you have to do and less formulas required to memorize.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you! Sorry I forgot to add the =1. Are my other values correct?
    $endgroup$
    – User231
    Dec 13 '18 at 16:36










  • $begingroup$
    The values appear correct to me!
    $endgroup$
    – Decaf-Math
    Dec 13 '18 at 16:39














3












3








3





$begingroup$

Based on the context that you provide, you're likely dealing with an ellipse whose equation is $${x^2over49} + {y^2over25} color{red}{= 1}.$$ The standard-form of an ellipse whose center is at the origin is given by $${x^2over a^2} + {y^2over b^2} = 1,$$ where $a$ and $b$ are the length of the horizontal and vertical axes respectively. Depending on whether $a$ is bigger or $b$ is bigger determines which one is the major axis. In this case, $a = 7$ and $b = 5$ by simple square root. We clearly see that $a = 7$ is bigger; since that is associated with the $x$ variable, you know that you have horizontal major axis, and thus you have the following vertices of the ellipse: $$(pm a, 0), (0,pm b) quad = quad (pm 7, 0), (pm 5, 0).$$ Geometrically speaking, this just means we go out $7$ both left and right from the center, and $5$ up and down from the center. Being able to interpret the ellipse geometrically in effect makes these kinds of problems much easier, as you can help reduce the number of formulas that you need to memorize.




This is especially true when you begin to shift the center to some general $(h,k)$; then the vertices become $(h pm a, k), (h, kpm b)$. This is a lot more "specific stuff" to remember, but if you interpret it geometrically, there is less math you have to do and less formulas required to memorize.






share|cite|improve this answer











$endgroup$



Based on the context that you provide, you're likely dealing with an ellipse whose equation is $${x^2over49} + {y^2over25} color{red}{= 1}.$$ The standard-form of an ellipse whose center is at the origin is given by $${x^2over a^2} + {y^2over b^2} = 1,$$ where $a$ and $b$ are the length of the horizontal and vertical axes respectively. Depending on whether $a$ is bigger or $b$ is bigger determines which one is the major axis. In this case, $a = 7$ and $b = 5$ by simple square root. We clearly see that $a = 7$ is bigger; since that is associated with the $x$ variable, you know that you have horizontal major axis, and thus you have the following vertices of the ellipse: $$(pm a, 0), (0,pm b) quad = quad (pm 7, 0), (pm 5, 0).$$ Geometrically speaking, this just means we go out $7$ both left and right from the center, and $5$ up and down from the center. Being able to interpret the ellipse geometrically in effect makes these kinds of problems much easier, as you can help reduce the number of formulas that you need to memorize.




This is especially true when you begin to shift the center to some general $(h,k)$; then the vertices become $(h pm a, k), (h, kpm b)$. This is a lot more "specific stuff" to remember, but if you interpret it geometrically, there is less math you have to do and less formulas required to memorize.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 13 '18 at 16:45

























answered Dec 13 '18 at 16:35









Decaf-MathDecaf-Math

3,422926




3,422926












  • $begingroup$
    Thank you! Sorry I forgot to add the =1. Are my other values correct?
    $endgroup$
    – User231
    Dec 13 '18 at 16:36










  • $begingroup$
    The values appear correct to me!
    $endgroup$
    – Decaf-Math
    Dec 13 '18 at 16:39


















  • $begingroup$
    Thank you! Sorry I forgot to add the =1. Are my other values correct?
    $endgroup$
    – User231
    Dec 13 '18 at 16:36










  • $begingroup$
    The values appear correct to me!
    $endgroup$
    – Decaf-Math
    Dec 13 '18 at 16:39
















$begingroup$
Thank you! Sorry I forgot to add the =1. Are my other values correct?
$endgroup$
– User231
Dec 13 '18 at 16:36




$begingroup$
Thank you! Sorry I forgot to add the =1. Are my other values correct?
$endgroup$
– User231
Dec 13 '18 at 16:36












$begingroup$
The values appear correct to me!
$endgroup$
– Decaf-Math
Dec 13 '18 at 16:39




$begingroup$
The values appear correct to me!
$endgroup$
– Decaf-Math
Dec 13 '18 at 16:39











1












$begingroup$

There's a lot going on here. First, you're missing an $=1$ in your equations.



The major axis is the longest "diameter" you can draw in the ellipse. The minor axis is the shortest. $a=7$ is what people would call the length of the major SEMI-axis.
So the length of the major axis is $14$.



It might help to plot the $x$ and $y$ intercepts (which you get by plugging in $y=0$ and $x=0$, respectively.) Those will be the endpoint of the axes you're looking for.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    So if the major axis is 7(2)=14, then the minor axis would be 5(2)=10?
    $endgroup$
    – User231
    Dec 13 '18 at 16:44










  • $begingroup$
    Yes. See mathopenref.com/ellipseaxes.html
    $endgroup$
    – B. Goddard
    Dec 13 '18 at 17:03
















1












$begingroup$

There's a lot going on here. First, you're missing an $=1$ in your equations.



The major axis is the longest "diameter" you can draw in the ellipse. The minor axis is the shortest. $a=7$ is what people would call the length of the major SEMI-axis.
So the length of the major axis is $14$.



It might help to plot the $x$ and $y$ intercepts (which you get by plugging in $y=0$ and $x=0$, respectively.) Those will be the endpoint of the axes you're looking for.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    So if the major axis is 7(2)=14, then the minor axis would be 5(2)=10?
    $endgroup$
    – User231
    Dec 13 '18 at 16:44










  • $begingroup$
    Yes. See mathopenref.com/ellipseaxes.html
    $endgroup$
    – B. Goddard
    Dec 13 '18 at 17:03














1












1








1





$begingroup$

There's a lot going on here. First, you're missing an $=1$ in your equations.



The major axis is the longest "diameter" you can draw in the ellipse. The minor axis is the shortest. $a=7$ is what people would call the length of the major SEMI-axis.
So the length of the major axis is $14$.



It might help to plot the $x$ and $y$ intercepts (which you get by plugging in $y=0$ and $x=0$, respectively.) Those will be the endpoint of the axes you're looking for.






share|cite|improve this answer









$endgroup$



There's a lot going on here. First, you're missing an $=1$ in your equations.



The major axis is the longest "diameter" you can draw in the ellipse. The minor axis is the shortest. $a=7$ is what people would call the length of the major SEMI-axis.
So the length of the major axis is $14$.



It might help to plot the $x$ and $y$ intercepts (which you get by plugging in $y=0$ and $x=0$, respectively.) Those will be the endpoint of the axes you're looking for.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 16:25









B. GoddardB. Goddard

20.2k21543




20.2k21543








  • 1




    $begingroup$
    So if the major axis is 7(2)=14, then the minor axis would be 5(2)=10?
    $endgroup$
    – User231
    Dec 13 '18 at 16:44










  • $begingroup$
    Yes. See mathopenref.com/ellipseaxes.html
    $endgroup$
    – B. Goddard
    Dec 13 '18 at 17:03














  • 1




    $begingroup$
    So if the major axis is 7(2)=14, then the minor axis would be 5(2)=10?
    $endgroup$
    – User231
    Dec 13 '18 at 16:44










  • $begingroup$
    Yes. See mathopenref.com/ellipseaxes.html
    $endgroup$
    – B. Goddard
    Dec 13 '18 at 17:03








1




1




$begingroup$
So if the major axis is 7(2)=14, then the minor axis would be 5(2)=10?
$endgroup$
– User231
Dec 13 '18 at 16:44




$begingroup$
So if the major axis is 7(2)=14, then the minor axis would be 5(2)=10?
$endgroup$
– User231
Dec 13 '18 at 16:44












$begingroup$
Yes. See mathopenref.com/ellipseaxes.html
$endgroup$
– B. Goddard
Dec 13 '18 at 17:03




$begingroup$
Yes. See mathopenref.com/ellipseaxes.html
$endgroup$
– B. Goddard
Dec 13 '18 at 17:03


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038207%2fhow-do-i-find-the-length-of-the-major-and-minor-axes-of-the-ellipse-fracx2%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
Read more