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Math for max and min x and y of an ellipse


Conversion from Standard Ellipse Function to General Ellipse FunctionNumber of ellipses to uniquely define a co-centered circumscribing ellipsePartial Integral of an ellipseArea of ellipse not in xy-planeFind foci and eccentricity of ellipse given either 5 points or its general equationWhy does the Ellipsograph/Trammel of Archimedes draw an ellipse, really?A new property of conics?1st and 2nd derivative test of max-minI got 5 ellipse parameters, 4 of which are coordinates of 2 focus points, what would the last one be? And how can I get the ellipse function?Parametrization of a rotated ellipse













2












$begingroup$


I see that the general equation of an ellipse centered at the origin is:



   (𝑎𝑥+𝑏𝑦)² + (𝑐𝑥+𝑑𝑦)² = 𝑟²



How do a, b, c, d relate to the coefficients of the general form?



   A𝑥² + B𝑥𝑦 + C𝑦² + D𝑥 + E𝑦 + F = 0



After some naive observation I believe:



   A = (𝑎² + 𝑐²)
   B = 2(𝑎𝑏 + 𝑐𝑑)
   C = (𝑏² + 𝑑²)
   F = -𝑟²



Pretty sure it's more complicated than that. The issue for me is to do the calculations based on numerical values for an ellipse in either standard or canonical form, as opposed to pencil and paper derivatives.










share|cite|improve this question







New contributor




TJ Sartain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    have an example $21x^2-6xy+29y^2+6x-58y+151=0$
    $endgroup$
    – jacky
    yesterday










  • $begingroup$
    OK, so using that example, F is an imaginary number. Also, I'm trying to go from the general form (with the capital letters) and get to a, b, c, d, and r.
    $endgroup$
    – TJ Sartain
    yesterday












  • $begingroup$
    I misrepresented my question from the very first line! Basically, I have an ellipse that is not necessarily centered at the origin and possibly rotated. I have either the standard form (or the canonical form as it is derivable) and I am trying to find the max and min x and y.
    $endgroup$
    – TJ Sartain
    yesterday
















2












$begingroup$


I see that the general equation of an ellipse centered at the origin is:



   (𝑎𝑥+𝑏𝑦)² + (𝑐𝑥+𝑑𝑦)² = 𝑟²



How do a, b, c, d relate to the coefficients of the general form?



   A𝑥² + B𝑥𝑦 + C𝑦² + D𝑥 + E𝑦 + F = 0



After some naive observation I believe:



   A = (𝑎² + 𝑐²)
   B = 2(𝑎𝑏 + 𝑐𝑑)
   C = (𝑏² + 𝑑²)
   F = -𝑟²



Pretty sure it's more complicated than that. The issue for me is to do the calculations based on numerical values for an ellipse in either standard or canonical form, as opposed to pencil and paper derivatives.










share|cite|improve this question







New contributor




TJ Sartain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    have an example $21x^2-6xy+29y^2+6x-58y+151=0$
    $endgroup$
    – jacky
    yesterday










  • $begingroup$
    OK, so using that example, F is an imaginary number. Also, I'm trying to go from the general form (with the capital letters) and get to a, b, c, d, and r.
    $endgroup$
    – TJ Sartain
    yesterday












  • $begingroup$
    I misrepresented my question from the very first line! Basically, I have an ellipse that is not necessarily centered at the origin and possibly rotated. I have either the standard form (or the canonical form as it is derivable) and I am trying to find the max and min x and y.
    $endgroup$
    – TJ Sartain
    yesterday














2












2








2





$begingroup$


I see that the general equation of an ellipse centered at the origin is:



   (𝑎𝑥+𝑏𝑦)² + (𝑐𝑥+𝑑𝑦)² = 𝑟²



How do a, b, c, d relate to the coefficients of the general form?



   A𝑥² + B𝑥𝑦 + C𝑦² + D𝑥 + E𝑦 + F = 0



After some naive observation I believe:



   A = (𝑎² + 𝑐²)
   B = 2(𝑎𝑏 + 𝑐𝑑)
   C = (𝑏² + 𝑑²)
   F = -𝑟²



Pretty sure it's more complicated than that. The issue for me is to do the calculations based on numerical values for an ellipse in either standard or canonical form, as opposed to pencil and paper derivatives.










share|cite|improve this question







New contributor




TJ Sartain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I see that the general equation of an ellipse centered at the origin is:



   (𝑎𝑥+𝑏𝑦)² + (𝑐𝑥+𝑑𝑦)² = 𝑟²



How do a, b, c, d relate to the coefficients of the general form?



   A𝑥² + B𝑥𝑦 + C𝑦² + D𝑥 + E𝑦 + F = 0



After some naive observation I believe:



   A = (𝑎² + 𝑐²)
   B = 2(𝑎𝑏 + 𝑐𝑑)
   C = (𝑏² + 𝑑²)
   F = -𝑟²



Pretty sure it's more complicated than that. The issue for me is to do the calculations based on numerical values for an ellipse in either standard or canonical form, as opposed to pencil and paper derivatives.







conic-sections maxima-minima






share|cite|improve this question







New contributor




TJ Sartain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




TJ Sartain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




TJ Sartain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked yesterday









TJ SartainTJ Sartain

111




111




New contributor




TJ Sartain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





TJ Sartain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






TJ Sartain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    have an example $21x^2-6xy+29y^2+6x-58y+151=0$
    $endgroup$
    – jacky
    yesterday










  • $begingroup$
    OK, so using that example, F is an imaginary number. Also, I'm trying to go from the general form (with the capital letters) and get to a, b, c, d, and r.
    $endgroup$
    – TJ Sartain
    yesterday












  • $begingroup$
    I misrepresented my question from the very first line! Basically, I have an ellipse that is not necessarily centered at the origin and possibly rotated. I have either the standard form (or the canonical form as it is derivable) and I am trying to find the max and min x and y.
    $endgroup$
    – TJ Sartain
    yesterday


















  • $begingroup$
    have an example $21x^2-6xy+29y^2+6x-58y+151=0$
    $endgroup$
    – jacky
    yesterday










  • $begingroup$
    OK, so using that example, F is an imaginary number. Also, I'm trying to go from the general form (with the capital letters) and get to a, b, c, d, and r.
    $endgroup$
    – TJ Sartain
    yesterday












  • $begingroup$
    I misrepresented my question from the very first line! Basically, I have an ellipse that is not necessarily centered at the origin and possibly rotated. I have either the standard form (or the canonical form as it is derivable) and I am trying to find the max and min x and y.
    $endgroup$
    – TJ Sartain
    yesterday
















$begingroup$
have an example $21x^2-6xy+29y^2+6x-58y+151=0$
$endgroup$
– jacky
yesterday




$begingroup$
have an example $21x^2-6xy+29y^2+6x-58y+151=0$
$endgroup$
– jacky
yesterday












$begingroup$
OK, so using that example, F is an imaginary number. Also, I'm trying to go from the general form (with the capital letters) and get to a, b, c, d, and r.
$endgroup$
– TJ Sartain
yesterday






$begingroup$
OK, so using that example, F is an imaginary number. Also, I'm trying to go from the general form (with the capital letters) and get to a, b, c, d, and r.
$endgroup$
– TJ Sartain
yesterday














$begingroup$
I misrepresented my question from the very first line! Basically, I have an ellipse that is not necessarily centered at the origin and possibly rotated. I have either the standard form (or the canonical form as it is derivable) and I am trying to find the max and min x and y.
$endgroup$
– TJ Sartain
yesterday




$begingroup$
I misrepresented my question from the very first line! Basically, I have an ellipse that is not necessarily centered at the origin and possibly rotated. I have either the standard form (or the canonical form as it is derivable) and I am trying to find the max and min x and y.
$endgroup$
– TJ Sartain
yesterday










1 Answer
1






active

oldest

votes


















1












$begingroup$

$$(ax+by)^2+(cx+dy)^2=r^2\
a^2x^2+2abxy+b^2y^2+c^2x^2+2cdxy+d^2y^2=r^2\
(a^2+c^2)x^2+(2ab+2cd)xy+(b^2+d^2)y^2-r^2=0\
therefore A=a^2+c^2B=2(ab+cd),C=(b^2+d^2),D=E=0,F=-r^2$$






share|cite|improve this answer








New contributor




hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$









  • 1




    $begingroup$
    This is what the OP wrote...
    $endgroup$
    – Peter Foreman
    yesterday










  • $begingroup$
    How do i convert it into $21x^2-6xy+29y^2+6x-58y+151=0$ standard form, I mean $pX^2+qY^2=1$ form
    $endgroup$
    – jacky
    yesterday












  • $begingroup$
    @PeterForeman The OP also said he thought it was more complicated than that. This answers the question, as I read it, and proves that it is not more complicated than that.
    $endgroup$
    – John Douma
    yesterday










  • $begingroup$
    Thanks, that IS how I got to the equivalencies I posted. But I have an equation where D and E are not 0. That's where I suspected there might be more to it. I would be surprised if I could simply ignore D and E but if someone actually verified that, I would accept it.
    $endgroup$
    – TJ Sartain
    yesterday










  • $begingroup$
    @TJSartain If $D$ and $E$ are not zero, then you therefore are dealing with something other than an ellipse centred at the origin.
    $endgroup$
    – Théophile
    yesterday











Your Answer





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1 Answer
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1 Answer
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active

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active

oldest

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active

oldest

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1












$begingroup$

$$(ax+by)^2+(cx+dy)^2=r^2\
a^2x^2+2abxy+b^2y^2+c^2x^2+2cdxy+d^2y^2=r^2\
(a^2+c^2)x^2+(2ab+2cd)xy+(b^2+d^2)y^2-r^2=0\
therefore A=a^2+c^2B=2(ab+cd),C=(b^2+d^2),D=E=0,F=-r^2$$






share|cite|improve this answer








New contributor




hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$









  • 1




    $begingroup$
    This is what the OP wrote...
    $endgroup$
    – Peter Foreman
    yesterday










  • $begingroup$
    How do i convert it into $21x^2-6xy+29y^2+6x-58y+151=0$ standard form, I mean $pX^2+qY^2=1$ form
    $endgroup$
    – jacky
    yesterday












  • $begingroup$
    @PeterForeman The OP also said he thought it was more complicated than that. This answers the question, as I read it, and proves that it is not more complicated than that.
    $endgroup$
    – John Douma
    yesterday










  • $begingroup$
    Thanks, that IS how I got to the equivalencies I posted. But I have an equation where D and E are not 0. That's where I suspected there might be more to it. I would be surprised if I could simply ignore D and E but if someone actually verified that, I would accept it.
    $endgroup$
    – TJ Sartain
    yesterday










  • $begingroup$
    @TJSartain If $D$ and $E$ are not zero, then you therefore are dealing with something other than an ellipse centred at the origin.
    $endgroup$
    – Théophile
    yesterday
















1












$begingroup$

$$(ax+by)^2+(cx+dy)^2=r^2\
a^2x^2+2abxy+b^2y^2+c^2x^2+2cdxy+d^2y^2=r^2\
(a^2+c^2)x^2+(2ab+2cd)xy+(b^2+d^2)y^2-r^2=0\
therefore A=a^2+c^2B=2(ab+cd),C=(b^2+d^2),D=E=0,F=-r^2$$






share|cite|improve this answer








New contributor




hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$









  • 1




    $begingroup$
    This is what the OP wrote...
    $endgroup$
    – Peter Foreman
    yesterday










  • $begingroup$
    How do i convert it into $21x^2-6xy+29y^2+6x-58y+151=0$ standard form, I mean $pX^2+qY^2=1$ form
    $endgroup$
    – jacky
    yesterday












  • $begingroup$
    @PeterForeman The OP also said he thought it was more complicated than that. This answers the question, as I read it, and proves that it is not more complicated than that.
    $endgroup$
    – John Douma
    yesterday










  • $begingroup$
    Thanks, that IS how I got to the equivalencies I posted. But I have an equation where D and E are not 0. That's where I suspected there might be more to it. I would be surprised if I could simply ignore D and E but if someone actually verified that, I would accept it.
    $endgroup$
    – TJ Sartain
    yesterday










  • $begingroup$
    @TJSartain If $D$ and $E$ are not zero, then you therefore are dealing with something other than an ellipse centred at the origin.
    $endgroup$
    – Théophile
    yesterday














1












1








1





$begingroup$

$$(ax+by)^2+(cx+dy)^2=r^2\
a^2x^2+2abxy+b^2y^2+c^2x^2+2cdxy+d^2y^2=r^2\
(a^2+c^2)x^2+(2ab+2cd)xy+(b^2+d^2)y^2-r^2=0\
therefore A=a^2+c^2B=2(ab+cd),C=(b^2+d^2),D=E=0,F=-r^2$$






share|cite|improve this answer








New contributor




hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$



$$(ax+by)^2+(cx+dy)^2=r^2\
a^2x^2+2abxy+b^2y^2+c^2x^2+2cdxy+d^2y^2=r^2\
(a^2+c^2)x^2+(2ab+2cd)xy+(b^2+d^2)y^2-r^2=0\
therefore A=a^2+c^2B=2(ab+cd),C=(b^2+d^2),D=E=0,F=-r^2$$







share|cite|improve this answer








New contributor




hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this answer



share|cite|improve this answer






New contributor




hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered yesterday









hokoxixehokoxixe

112




112




New contributor




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New contributor





hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    This is what the OP wrote...
    $endgroup$
    – Peter Foreman
    yesterday










  • $begingroup$
    How do i convert it into $21x^2-6xy+29y^2+6x-58y+151=0$ standard form, I mean $pX^2+qY^2=1$ form
    $endgroup$
    – jacky
    yesterday












  • $begingroup$
    @PeterForeman The OP also said he thought it was more complicated than that. This answers the question, as I read it, and proves that it is not more complicated than that.
    $endgroup$
    – John Douma
    yesterday










  • $begingroup$
    Thanks, that IS how I got to the equivalencies I posted. But I have an equation where D and E are not 0. That's where I suspected there might be more to it. I would be surprised if I could simply ignore D and E but if someone actually verified that, I would accept it.
    $endgroup$
    – TJ Sartain
    yesterday










  • $begingroup$
    @TJSartain If $D$ and $E$ are not zero, then you therefore are dealing with something other than an ellipse centred at the origin.
    $endgroup$
    – Théophile
    yesterday














  • 1




    $begingroup$
    This is what the OP wrote...
    $endgroup$
    – Peter Foreman
    yesterday










  • $begingroup$
    How do i convert it into $21x^2-6xy+29y^2+6x-58y+151=0$ standard form, I mean $pX^2+qY^2=1$ form
    $endgroup$
    – jacky
    yesterday












  • $begingroup$
    @PeterForeman The OP also said he thought it was more complicated than that. This answers the question, as I read it, and proves that it is not more complicated than that.
    $endgroup$
    – John Douma
    yesterday










  • $begingroup$
    Thanks, that IS how I got to the equivalencies I posted. But I have an equation where D and E are not 0. That's where I suspected there might be more to it. I would be surprised if I could simply ignore D and E but if someone actually verified that, I would accept it.
    $endgroup$
    – TJ Sartain
    yesterday










  • $begingroup$
    @TJSartain If $D$ and $E$ are not zero, then you therefore are dealing with something other than an ellipse centred at the origin.
    $endgroup$
    – Théophile
    yesterday








1




1




$begingroup$
This is what the OP wrote...
$endgroup$
– Peter Foreman
yesterday




$begingroup$
This is what the OP wrote...
$endgroup$
– Peter Foreman
yesterday












$begingroup$
How do i convert it into $21x^2-6xy+29y^2+6x-58y+151=0$ standard form, I mean $pX^2+qY^2=1$ form
$endgroup$
– jacky
yesterday






$begingroup$
How do i convert it into $21x^2-6xy+29y^2+6x-58y+151=0$ standard form, I mean $pX^2+qY^2=1$ form
$endgroup$
– jacky
yesterday














$begingroup$
@PeterForeman The OP also said he thought it was more complicated than that. This answers the question, as I read it, and proves that it is not more complicated than that.
$endgroup$
– John Douma
yesterday




$begingroup$
@PeterForeman The OP also said he thought it was more complicated than that. This answers the question, as I read it, and proves that it is not more complicated than that.
$endgroup$
– John Douma
yesterday












$begingroup$
Thanks, that IS how I got to the equivalencies I posted. But I have an equation where D and E are not 0. That's where I suspected there might be more to it. I would be surprised if I could simply ignore D and E but if someone actually verified that, I would accept it.
$endgroup$
– TJ Sartain
yesterday




$begingroup$
Thanks, that IS how I got to the equivalencies I posted. But I have an equation where D and E are not 0. That's where I suspected there might be more to it. I would be surprised if I could simply ignore D and E but if someone actually verified that, I would accept it.
$endgroup$
– TJ Sartain
yesterday












$begingroup$
@TJSartain If $D$ and $E$ are not zero, then you therefore are dealing with something other than an ellipse centred at the origin.
$endgroup$
– Théophile
yesterday




$begingroup$
@TJSartain If $D$ and $E$ are not zero, then you therefore are dealing with something other than an ellipse centred at the origin.
$endgroup$
– Théophile
yesterday










TJ Sartain is a new contributor. Be nice, and check out our Code of Conduct.










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TJ Sartain is a new contributor. Be nice, and check out our Code of Conduct.












TJ Sartain is a new contributor. Be nice, and check out our Code of Conduct.
















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