Finding equation for conic section given five points












2












$begingroup$


Problem:



Given the points $$(0,1),(0,-1),(2,0),(-2,0),(1,1)$$ find the equation for the conic section that passes through these points.



My attempt:



Using the general equation for a conic section, $$ax^2+bxy+cy^2+dx+ey+f=0$$ I inserted the points to make a system of 5 equations, which was reduced to the following, with $a$ as a free variable.



$$b=-a \c=4a \d=e=0 \f=-4a$$



Putting this into the general equation for a conic, I got $$ax^2-axy+4ay^2-4a=0 (1)$$



My question:



Can I determine which type of conic section this is? I read somewhere that I can determine it by finding out the sign of $b^2+4ac$, where I ended up with $b^2-4ac = -15a^2 < 0$ which is supposed to be an ellipse. However, I can't find my way back to the place where I read this. Looking at the points on a graph, it seems visually obvious that this is an ellipse. However, I can't prove it unless the mentioned method is valid.



If it really is an ellipse, how can I algebraically manipulate $(1)$ into the form most known for ellipses?






conic-sections







share|cite|improve this question









$endgroup$












  • $begingroup$
    The form most known for ellipses has major and minor axes parallel to the $x$ and $y$ axes, and thus no cross term (i.e. $b = 0$). Without rotating the coordinate system there is no hope of writing this as $frac{(x-i)^2}{j} + frac{(y-k)^2}{l} = 1$
    $endgroup$
    – Arthur
    Jan 10 '15 at 1:52










  • $begingroup$
    @Amzoti - Yes, there is the $b^2-4ac$ bit I was talking about. Using that calculator confirmed what I thought, but I still need to prove that $x^2-xy+4y^2-4=0$ represents an ellipse. Also, any non-zero $a$ will yield the same ellipse. This is obvious since you can factor it out from $(1)$.
    $endgroup$
    – Alec
    Jan 10 '15 at 2:14










  • $begingroup$
    @Arthur - You're right. Drawing the equation in Geogebra shows a slightly rotated ellipse. Could $(1)$ after factoring and striking $a neq 0$ be the answer to the problem?
    $endgroup$
    – Alec
    Jan 10 '15 at 2:16










  • $begingroup$
    Why not to use the method described in en.wikipedia.org/wiki/…
    $endgroup$
    – Claude Leibovici
    Jan 10 '15 at 4:23










  • $begingroup$
    For finding the equation in the first place, solving is the system requires a guess about which variable to leave "free"; and, of course, there's all that solving work to do. Using the determinant mentioned in this answer, one generates the equation directly ... albeit with the work of expanding a $6times 6$ determinant.
    $endgroup$
    – Blue
    Aug 18 '18 at 16:04
















2












$begingroup$


Problem:



Given the points $$(0,1),(0,-1),(2,0),(-2,0),(1,1)$$ find the equation for the conic section that passes through these points.



My attempt:



Using the general equation for a conic section, $$ax^2+bxy+cy^2+dx+ey+f=0$$ I inserted the points to make a system of 5 equations, which was reduced to the following, with $a$ as a free variable.



$$b=-a \c=4a \d=e=0 \f=-4a$$



Putting this into the general equation for a conic, I got $$ax^2-axy+4ay^2-4a=0 (1)$$



My question:



Can I determine which type of conic section this is? I read somewhere that I can determine it by finding out the sign of $b^2+4ac$, where I ended up with $b^2-4ac = -15a^2 < 0$ which is supposed to be an ellipse. However, I can't find my way back to the place where I read this. Looking at the points on a graph, it seems visually obvious that this is an ellipse. However, I can't prove it unless the mentioned method is valid.



If it really is an ellipse, how can I algebraically manipulate $(1)$ into the form most known for ellipses?






conic-sections







share|cite|improve this question









$endgroup$












  • $begingroup$
    The form most known for ellipses has major and minor axes parallel to the $x$ and $y$ axes, and thus no cross term (i.e. $b = 0$). Without rotating the coordinate system there is no hope of writing this as $frac{(x-i)^2}{j} + frac{(y-k)^2}{l} = 1$
    $endgroup$
    – Arthur
    Jan 10 '15 at 1:52










  • $begingroup$
    @Amzoti - Yes, there is the $b^2-4ac$ bit I was talking about. Using that calculator confirmed what I thought, but I still need to prove that $x^2-xy+4y^2-4=0$ represents an ellipse. Also, any non-zero $a$ will yield the same ellipse. This is obvious since you can factor it out from $(1)$.
    $endgroup$
    – Alec
    Jan 10 '15 at 2:14










  • $begingroup$
    @Arthur - You're right. Drawing the equation in Geogebra shows a slightly rotated ellipse. Could $(1)$ after factoring and striking $a neq 0$ be the answer to the problem?
    $endgroup$
    – Alec
    Jan 10 '15 at 2:16










  • $begingroup$
    Why not to use the method described in en.wikipedia.org/wiki/…
    $endgroup$
    – Claude Leibovici
    Jan 10 '15 at 4:23










  • $begingroup$
    For finding the equation in the first place, solving is the system requires a guess about which variable to leave "free"; and, of course, there's all that solving work to do. Using the determinant mentioned in this answer, one generates the equation directly ... albeit with the work of expanding a $6times 6$ determinant.
    $endgroup$
    – Blue
    Aug 18 '18 at 16:04














2












2








2


2



$begingroup$


Problem:



Given the points $$(0,1),(0,-1),(2,0),(-2,0),(1,1)$$ find the equation for the conic section that passes through these points.



My attempt:



Using the general equation for a conic section, $$ax^2+bxy+cy^2+dx+ey+f=0$$ I inserted the points to make a system of 5 equations, which was reduced to the following, with $a$ as a free variable.



$$b=-a \c=4a \d=e=0 \f=-4a$$



Putting this into the general equation for a conic, I got $$ax^2-axy+4ay^2-4a=0 (1)$$



My question:



Can I determine which type of conic section this is? I read somewhere that I can determine it by finding out the sign of $b^2+4ac$, where I ended up with $b^2-4ac = -15a^2 < 0$ which is supposed to be an ellipse. However, I can't find my way back to the place where I read this. Looking at the points on a graph, it seems visually obvious that this is an ellipse. However, I can't prove it unless the mentioned method is valid.



If it really is an ellipse, how can I algebraically manipulate $(1)$ into the form most known for ellipses?






conic-sections







share|cite|improve this question









$endgroup$




Problem:



Given the points $$(0,1),(0,-1),(2,0),(-2,0),(1,1)$$ find the equation for the conic section that passes through these points.



My attempt:



Using the general equation for a conic section, $$ax^2+bxy+cy^2+dx+ey+f=0$$ I inserted the points to make a system of 5 equations, which was reduced to the following, with $a$ as a free variable.



$$b=-a \c=4a \d=e=0 \f=-4a$$



Putting this into the general equation for a conic, I got $$ax^2-axy+4ay^2-4a=0 (1)$$



My question:



Can I determine which type of conic section this is? I read somewhere that I can determine it by finding out the sign of $b^2+4ac$, where I ended up with $b^2-4ac = -15a^2 < 0$ which is supposed to be an ellipse. However, I can't find my way back to the place where I read this. Looking at the points on a graph, it seems visually obvious that this is an ellipse. However, I can't prove it unless the mentioned method is valid.



If it really is an ellipse, how can I algebraically manipulate $(1)$ into the form most known for ellipses?






conic-sections




conic-sections






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 10 '15 at 1:37









AlecAlec

2,20211539




2,20211539












  • $begingroup$
    The form most known for ellipses has major and minor axes parallel to the $x$ and $y$ axes, and thus no cross term (i.e. $b = 0$). Without rotating the coordinate system there is no hope of writing this as $frac{(x-i)^2}{j} + frac{(y-k)^2}{l} = 1$
    $endgroup$
    – Arthur
    Jan 10 '15 at 1:52










  • $begingroup$
    @Amzoti - Yes, there is the $b^2-4ac$ bit I was talking about. Using that calculator confirmed what I thought, but I still need to prove that $x^2-xy+4y^2-4=0$ represents an ellipse. Also, any non-zero $a$ will yield the same ellipse. This is obvious since you can factor it out from $(1)$.
    $endgroup$
    – Alec
    Jan 10 '15 at 2:14










  • $begingroup$
    @Arthur - You're right. Drawing the equation in Geogebra shows a slightly rotated ellipse. Could $(1)$ after factoring and striking $a neq 0$ be the answer to the problem?
    $endgroup$
    – Alec
    Jan 10 '15 at 2:16










  • $begingroup$
    Why not to use the method described in en.wikipedia.org/wiki/…
    $endgroup$
    – Claude Leibovici
    Jan 10 '15 at 4:23










  • $begingroup$
    For finding the equation in the first place, solving is the system requires a guess about which variable to leave "free"; and, of course, there's all that solving work to do. Using the determinant mentioned in this answer, one generates the equation directly ... albeit with the work of expanding a $6times 6$ determinant.
    $endgroup$
    – Blue
    Aug 18 '18 at 16:04


















  • $begingroup$
    The form most known for ellipses has major and minor axes parallel to the $x$ and $y$ axes, and thus no cross term (i.e. $b = 0$). Without rotating the coordinate system there is no hope of writing this as $frac{(x-i)^2}{j} + frac{(y-k)^2}{l} = 1$
    $endgroup$
    – Arthur
    Jan 10 '15 at 1:52










  • $begingroup$
    @Amzoti - Yes, there is the $b^2-4ac$ bit I was talking about. Using that calculator confirmed what I thought, but I still need to prove that $x^2-xy+4y^2-4=0$ represents an ellipse. Also, any non-zero $a$ will yield the same ellipse. This is obvious since you can factor it out from $(1)$.
    $endgroup$
    – Alec
    Jan 10 '15 at 2:14










  • $begingroup$
    @Arthur - You're right. Drawing the equation in Geogebra shows a slightly rotated ellipse. Could $(1)$ after factoring and striking $a neq 0$ be the answer to the problem?
    $endgroup$
    – Alec
    Jan 10 '15 at 2:16










  • $begingroup$
    Why not to use the method described in en.wikipedia.org/wiki/…
    $endgroup$
    – Claude Leibovici
    Jan 10 '15 at 4:23










  • $begingroup$
    For finding the equation in the first place, solving is the system requires a guess about which variable to leave "free"; and, of course, there's all that solving work to do. Using the determinant mentioned in this answer, one generates the equation directly ... albeit with the work of expanding a $6times 6$ determinant.
    $endgroup$
    – Blue
    Aug 18 '18 at 16:04
















$begingroup$
The form most known for ellipses has major and minor axes parallel to the $x$ and $y$ axes, and thus no cross term (i.e. $b = 0$). Without rotating the coordinate system there is no hope of writing this as $frac{(x-i)^2}{j} + frac{(y-k)^2}{l} = 1$
$endgroup$
– Arthur
Jan 10 '15 at 1:52




$begingroup$
The form most known for ellipses has major and minor axes parallel to the $x$ and $y$ axes, and thus no cross term (i.e. $b = 0$). Without rotating the coordinate system there is no hope of writing this as $frac{(x-i)^2}{j} + frac{(y-k)^2}{l} = 1$
$endgroup$
– Arthur
Jan 10 '15 at 1:52












$begingroup$
@Amzoti - Yes, there is the $b^2-4ac$ bit I was talking about. Using that calculator confirmed what I thought, but I still need to prove that $x^2-xy+4y^2-4=0$ represents an ellipse. Also, any non-zero $a$ will yield the same ellipse. This is obvious since you can factor it out from $(1)$.
$endgroup$
– Alec
Jan 10 '15 at 2:14




$begingroup$
@Amzoti - Yes, there is the $b^2-4ac$ bit I was talking about. Using that calculator confirmed what I thought, but I still need to prove that $x^2-xy+4y^2-4=0$ represents an ellipse. Also, any non-zero $a$ will yield the same ellipse. This is obvious since you can factor it out from $(1)$.
$endgroup$
– Alec
Jan 10 '15 at 2:14












$begingroup$
@Arthur - You're right. Drawing the equation in Geogebra shows a slightly rotated ellipse. Could $(1)$ after factoring and striking $a neq 0$ be the answer to the problem?
$endgroup$
– Alec
Jan 10 '15 at 2:16




$begingroup$
@Arthur - You're right. Drawing the equation in Geogebra shows a slightly rotated ellipse. Could $(1)$ after factoring and striking $a neq 0$ be the answer to the problem?
$endgroup$
– Alec
Jan 10 '15 at 2:16












$begingroup$
Why not to use the method described in en.wikipedia.org/wiki/…
$endgroup$
– Claude Leibovici
Jan 10 '15 at 4:23




$begingroup$
Why not to use the method described in en.wikipedia.org/wiki/…
$endgroup$
– Claude Leibovici
Jan 10 '15 at 4:23












$begingroup$
For finding the equation in the first place, solving is the system requires a guess about which variable to leave "free"; and, of course, there's all that solving work to do. Using the determinant mentioned in this answer, one generates the equation directly ... albeit with the work of expanding a $6times 6$ determinant.
$endgroup$
– Blue
Aug 18 '18 at 16:04




$begingroup$
For finding the equation in the first place, solving is the system requires a guess about which variable to leave "free"; and, of course, there's all that solving work to do. Using the determinant mentioned in this answer, one generates the equation directly ... albeit with the work of expanding a $6times 6$ determinant.
$endgroup$
– Blue
Aug 18 '18 at 16:04










1 Answer
1






active

oldest

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0












$begingroup$


Can I determine which type of conic section this is? I read somewhere that I can determine it by finding out the sign of b^2+4ac, where I ended up with b^2−4ac=−15a^2<0 which is supposed to be an ellipse.




Yes, it is b^2-4ac actually.



First, solve the determinant



$begin{bmatrix}a & b/2 & d/2\b/2 & c & e/2\d/2 & e/2 & fend{bmatrix}$ = p



Now, if:-




  1. p=0 and b^2-4ac>0, it is a pair of distinct straight lines.

  2. p=0 and b^2-4ac=0, it is a either the same line or pair of parrallel lines.

  3. p=0 and b^2-4ac<0, it is a pair of imaginary lines.

  4. p≠0 and b^2-4ac=0, it is a parabola.

  5. p≠0 and b^2-4ac<0, it is an ellipse.

  6. p≠0 and b^2-4ac>0, it is a hyperbola.

  7. p≠0, b=0 and a=c, it is a circle.


Sorry, but I'm not that good with MathJax.






share|cite|improve this answer









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    $begingroup$


    Can I determine which type of conic section this is? I read somewhere that I can determine it by finding out the sign of b^2+4ac, where I ended up with b^2−4ac=−15a^2<0 which is supposed to be an ellipse.




    Yes, it is b^2-4ac actually.



    First, solve the determinant



    $begin{bmatrix}a & b/2 & d/2\b/2 & c & e/2\d/2 & e/2 & fend{bmatrix}$ = p



    Now, if:-




    1. p=0 and b^2-4ac>0, it is a pair of distinct straight lines.

    2. p=0 and b^2-4ac=0, it is a either the same line or pair of parrallel lines.

    3. p=0 and b^2-4ac<0, it is a pair of imaginary lines.

    4. p≠0 and b^2-4ac=0, it is a parabola.

    5. p≠0 and b^2-4ac<0, it is an ellipse.

    6. p≠0 and b^2-4ac>0, it is a hyperbola.

    7. p≠0, b=0 and a=c, it is a circle.


    Sorry, but I'm not that good with MathJax.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$


      Can I determine which type of conic section this is? I read somewhere that I can determine it by finding out the sign of b^2+4ac, where I ended up with b^2−4ac=−15a^2<0 which is supposed to be an ellipse.




      Yes, it is b^2-4ac actually.



      First, solve the determinant



      $begin{bmatrix}a & b/2 & d/2\b/2 & c & e/2\d/2 & e/2 & fend{bmatrix}$ = p



      Now, if:-




      1. p=0 and b^2-4ac>0, it is a pair of distinct straight lines.

      2. p=0 and b^2-4ac=0, it is a either the same line or pair of parrallel lines.

      3. p=0 and b^2-4ac<0, it is a pair of imaginary lines.

      4. p≠0 and b^2-4ac=0, it is a parabola.

      5. p≠0 and b^2-4ac<0, it is an ellipse.

      6. p≠0 and b^2-4ac>0, it is a hyperbola.

      7. p≠0, b=0 and a=c, it is a circle.


      Sorry, but I'm not that good with MathJax.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$


        Can I determine which type of conic section this is? I read somewhere that I can determine it by finding out the sign of b^2+4ac, where I ended up with b^2−4ac=−15a^2<0 which is supposed to be an ellipse.




        Yes, it is b^2-4ac actually.



        First, solve the determinant



        $begin{bmatrix}a & b/2 & d/2\b/2 & c & e/2\d/2 & e/2 & fend{bmatrix}$ = p



        Now, if:-




        1. p=0 and b^2-4ac>0, it is a pair of distinct straight lines.

        2. p=0 and b^2-4ac=0, it is a either the same line or pair of parrallel lines.

        3. p=0 and b^2-4ac<0, it is a pair of imaginary lines.

        4. p≠0 and b^2-4ac=0, it is a parabola.

        5. p≠0 and b^2-4ac<0, it is an ellipse.

        6. p≠0 and b^2-4ac>0, it is a hyperbola.

        7. p≠0, b=0 and a=c, it is a circle.


        Sorry, but I'm not that good with MathJax.






        share|cite|improve this answer









        $endgroup$




        Can I determine which type of conic section this is? I read somewhere that I can determine it by finding out the sign of b^2+4ac, where I ended up with b^2−4ac=−15a^2<0 which is supposed to be an ellipse.




        Yes, it is b^2-4ac actually.



        First, solve the determinant



        $begin{bmatrix}a & b/2 & d/2\b/2 & c & e/2\d/2 & e/2 & fend{bmatrix}$ = p



        Now, if:-




        1. p=0 and b^2-4ac>0, it is a pair of distinct straight lines.

        2. p=0 and b^2-4ac=0, it is a either the same line or pair of parrallel lines.

        3. p=0 and b^2-4ac<0, it is a pair of imaginary lines.

        4. p≠0 and b^2-4ac=0, it is a parabola.

        5. p≠0 and b^2-4ac<0, it is an ellipse.

        6. p≠0 and b^2-4ac>0, it is a hyperbola.

        7. p≠0, b=0 and a=c, it is a circle.


        Sorry, but I'm not that good with MathJax.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 18 '18 at 15:54









        user79161user79161

        191310




        191310






























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            Dieringhausen

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            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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