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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$begingroup$


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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            Marschland

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            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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