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







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$endgroup$








  • 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












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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


















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