Intersection of a cylinder and a plane











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0
down vote

favorite












Given a plane:
$$x - y + z = 0$$



And a cylinder
$$x^2 + y^2 = 2$$



Why can't I get the intersection of the two by equaling both equations? i.e.
$$x^2 + y^2 - 2 = x - y + z implies x^2+y^2 -x + y - z -2=0$$



Instead, I have to isolate $x$ for example in eq(1) and substitute in eq(2).










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  • You have got it. You are not looking for a triple of numbers $(a,b,c)$; you are looking for a curve. You can complete the squares on the left hand side to get it into a more canonical form but that is the solution.
    – John Douma
    Nov 23 at 15:57










  • $x^2 + y^2 = 2$ is cylinder with axis in the $z$ direction. Take $z=0$ giving a slice of the cylinder and the plane $x - y + z = 0$ on the $x-y$ plane.
    – Yadati Kiran
    Nov 23 at 16:00















up vote
0
down vote

favorite












Given a plane:
$$x - y + z = 0$$



And a cylinder
$$x^2 + y^2 = 2$$



Why can't I get the intersection of the two by equaling both equations? i.e.
$$x^2 + y^2 - 2 = x - y + z implies x^2+y^2 -x + y - z -2=0$$



Instead, I have to isolate $x$ for example in eq(1) and substitute in eq(2).










share|cite|improve this question






















  • You have got it. You are not looking for a triple of numbers $(a,b,c)$; you are looking for a curve. You can complete the squares on the left hand side to get it into a more canonical form but that is the solution.
    – John Douma
    Nov 23 at 15:57










  • $x^2 + y^2 = 2$ is cylinder with axis in the $z$ direction. Take $z=0$ giving a slice of the cylinder and the plane $x - y + z = 0$ on the $x-y$ plane.
    – Yadati Kiran
    Nov 23 at 16:00













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Given a plane:
$$x - y + z = 0$$



And a cylinder
$$x^2 + y^2 = 2$$



Why can't I get the intersection of the two by equaling both equations? i.e.
$$x^2 + y^2 - 2 = x - y + z implies x^2+y^2 -x + y - z -2=0$$



Instead, I have to isolate $x$ for example in eq(1) and substitute in eq(2).










share|cite|improve this question













Given a plane:
$$x - y + z = 0$$



And a cylinder
$$x^2 + y^2 = 2$$



Why can't I get the intersection of the two by equaling both equations? i.e.
$$x^2 + y^2 - 2 = x - y + z implies x^2+y^2 -x + y - z -2=0$$



Instead, I have to isolate $x$ for example in eq(1) and substitute in eq(2).







geometry plane-geometry






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asked Nov 23 at 15:52









Ivan

1083




1083












  • You have got it. You are not looking for a triple of numbers $(a,b,c)$; you are looking for a curve. You can complete the squares on the left hand side to get it into a more canonical form but that is the solution.
    – John Douma
    Nov 23 at 15:57










  • $x^2 + y^2 = 2$ is cylinder with axis in the $z$ direction. Take $z=0$ giving a slice of the cylinder and the plane $x - y + z = 0$ on the $x-y$ plane.
    – Yadati Kiran
    Nov 23 at 16:00


















  • You have got it. You are not looking for a triple of numbers $(a,b,c)$; you are looking for a curve. You can complete the squares on the left hand side to get it into a more canonical form but that is the solution.
    – John Douma
    Nov 23 at 15:57










  • $x^2 + y^2 = 2$ is cylinder with axis in the $z$ direction. Take $z=0$ giving a slice of the cylinder and the plane $x - y + z = 0$ on the $x-y$ plane.
    – Yadati Kiran
    Nov 23 at 16:00
















You have got it. You are not looking for a triple of numbers $(a,b,c)$; you are looking for a curve. You can complete the squares on the left hand side to get it into a more canonical form but that is the solution.
– John Douma
Nov 23 at 15:57




You have got it. You are not looking for a triple of numbers $(a,b,c)$; you are looking for a curve. You can complete the squares on the left hand side to get it into a more canonical form but that is the solution.
– John Douma
Nov 23 at 15:57












$x^2 + y^2 = 2$ is cylinder with axis in the $z$ direction. Take $z=0$ giving a slice of the cylinder and the plane $x - y + z = 0$ on the $x-y$ plane.
– Yadati Kiran
Nov 23 at 16:00




$x^2 + y^2 = 2$ is cylinder with axis in the $z$ direction. Take $z=0$ giving a slice of the cylinder and the plane $x - y + z = 0$ on the $x-y$ plane.
– Yadati Kiran
Nov 23 at 16:00










5 Answers
5






active

oldest

votes

















up vote
1
down vote













Generally, a curve in 3-space can't be expressed with one equation. Try at 2-D example. Take the two equations



$$x^2+y^2 =1$$



and



$$x=0$$.



The intersection is just the two points $(0,pm 1)$. But if we do what you did above, we have



$$x^2+y^2=1 =x$$



which has infinitely many solutions.






share|cite|improve this answer




























    up vote
    1
    down vote













    The intesection is an ellipse in the space and the equantion is defined by the two curves, that is




    • $z=y-x$

    • $x^2+y^2=2$


    As an alternative rpresntation we can use




    • $x=sqrt 2 cos t$

    • $y=sqrt 2 sin t$

    • $z=sqrt 2 (sin t-cos t)$


    with $tin[0,2pi)$.






    share|cite|improve this answer




























      up vote
      0
      down vote













      To elaborate on why the technique you used in general doesn't work, it is because you introduce a new degree of freedom.



      More precisely, given two equations
      $$f(x,y,z)=0,$$ and $$g(x,y,z)=0,$$ if you consider the equation $$f(x,y,z)=g(x,y,z),$$ the solutions to this new equation should contain the entire intersection of the two original surfaces, but it will also contain the imtersection of the surfaces $$f(x,y,z)=1,$$ and $$g(x,y,z)=1.$$ Or, in other words, the equation $$f(x,y,z)=g(x,y,z)$$ can have solutions where $$f(x,y,z)neq 0,$$ but these solutions violate the original equations.






      share|cite|improve this answer




























        up vote
        0
        down vote













        I think that you have been badly misled by sloppy teaching at some level, probably high-school. You should never expect that you can find the common solution(s) of two equations by setting them equal, except in some rather special situations. Let me go into some detail:



        One case where your strategy will work is when you have two equations, $y=f(x)$ and $y=g(x)$, and want to know where the two graphs intersect. Then you may (and should) write $f(x)=g(x)$ and solve the single equation $f(x)-g(x)=0$ for $x$. The reason this works is that by setting $f=g$, you have in effect eliminated the letter $y$ from the picture, and can now use one-variable methods to come to the desired answer.



        Your particular problem presented you with a three-variable situation, and, as @B.Goddard and @gimusi have said in their answers, the simple methods applicable to two-variable problems have to be handled rather differently.



        But let me come back to the two-variable case, where you expect finitely many common points to two planar curves. There, setting $f=g$ in the case that you were intersecting two graphs was just a neat and easy trick for eliminating $y$ from the situation, and in the two-variable situation, you should always base your strategy on eliminating one of the variables. Take an interesting case like intersecting a parabola with a circle, like $y=x^2+x-2$ and $x^2+y^2=4$. Here you easily eliminate $y$ by substituting its expression in the first equation for the appearance(s) of $y$ in the second, to get
        begin{align}
        x^2+(x^2+x-2)^2&=4\
        x^2+x^4+2x^3-3x^2-4x+4&=4\
        x^4+2x^3-2x^2-4x&=0\
        x(x+2)(x^2-2)&=0,,
        end{align}

        to give the four(*) points $(-2,0)$, $(-sqrt2,-sqrt2)$, $(0,-2)$, and $(-sqrt2,-sqrt2)$ of the intersection.



        In cases where neither equation is already solved for
        $x$ nor $y$, you may have to try to solve one by your own means. If the equations are of higher degree than $2$, this can get tricky.



        (*) According to a wonderful theorem of Bézout, any two different conic sections will have exactly four points in their intersection, as long as you count properly.






        share|cite|improve this answer




























          up vote
          0
          down vote













          Eliminating $x$ we get $(y-z)$ orthogonal plane projection ( and eliminating $y$ gives $(x-y)$ plane projection) etc.



          A parameter can be chosen to represent the three coordinates of the curve of intersection by eliminating or isolating one variable.



          But by directly adding the two equations together in linear combinations as in an example



          $$ lambda(x^2+y^2) + (lambda-1) (x - y + z +2) =0 $$



          we get a set or family of second degree conicoids/quadrics passing through their common intersection curve ( ellipse in this case). For equal weightage as in the case $lambda=frac12$ you mention, it is a paraboloid.






          share|cite|improve this answer























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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

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            active

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            up vote
            1
            down vote













            Generally, a curve in 3-space can't be expressed with one equation. Try at 2-D example. Take the two equations



            $$x^2+y^2 =1$$



            and



            $$x=0$$.



            The intersection is just the two points $(0,pm 1)$. But if we do what you did above, we have



            $$x^2+y^2=1 =x$$



            which has infinitely many solutions.






            share|cite|improve this answer

























              up vote
              1
              down vote













              Generally, a curve in 3-space can't be expressed with one equation. Try at 2-D example. Take the two equations



              $$x^2+y^2 =1$$



              and



              $$x=0$$.



              The intersection is just the two points $(0,pm 1)$. But if we do what you did above, we have



              $$x^2+y^2=1 =x$$



              which has infinitely many solutions.






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                Generally, a curve in 3-space can't be expressed with one equation. Try at 2-D example. Take the two equations



                $$x^2+y^2 =1$$



                and



                $$x=0$$.



                The intersection is just the two points $(0,pm 1)$. But if we do what you did above, we have



                $$x^2+y^2=1 =x$$



                which has infinitely many solutions.






                share|cite|improve this answer












                Generally, a curve in 3-space can't be expressed with one equation. Try at 2-D example. Take the two equations



                $$x^2+y^2 =1$$



                and



                $$x=0$$.



                The intersection is just the two points $(0,pm 1)$. But if we do what you did above, we have



                $$x^2+y^2=1 =x$$



                which has infinitely many solutions.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 23 at 16:00









                B. Goddard

                18.2k21340




                18.2k21340






















                    up vote
                    1
                    down vote













                    The intesection is an ellipse in the space and the equantion is defined by the two curves, that is




                    • $z=y-x$

                    • $x^2+y^2=2$


                    As an alternative rpresntation we can use




                    • $x=sqrt 2 cos t$

                    • $y=sqrt 2 sin t$

                    • $z=sqrt 2 (sin t-cos t)$


                    with $tin[0,2pi)$.






                    share|cite|improve this answer

























                      up vote
                      1
                      down vote













                      The intesection is an ellipse in the space and the equantion is defined by the two curves, that is




                      • $z=y-x$

                      • $x^2+y^2=2$


                      As an alternative rpresntation we can use




                      • $x=sqrt 2 cos t$

                      • $y=sqrt 2 sin t$

                      • $z=sqrt 2 (sin t-cos t)$


                      with $tin[0,2pi)$.






                      share|cite|improve this answer























                        up vote
                        1
                        down vote










                        up vote
                        1
                        down vote









                        The intesection is an ellipse in the space and the equantion is defined by the two curves, that is




                        • $z=y-x$

                        • $x^2+y^2=2$


                        As an alternative rpresntation we can use




                        • $x=sqrt 2 cos t$

                        • $y=sqrt 2 sin t$

                        • $z=sqrt 2 (sin t-cos t)$


                        with $tin[0,2pi)$.






                        share|cite|improve this answer












                        The intesection is an ellipse in the space and the equantion is defined by the two curves, that is




                        • $z=y-x$

                        • $x^2+y^2=2$


                        As an alternative rpresntation we can use




                        • $x=sqrt 2 cos t$

                        • $y=sqrt 2 sin t$

                        • $z=sqrt 2 (sin t-cos t)$


                        with $tin[0,2pi)$.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Nov 23 at 16:00









                        gimusi

                        88.5k74394




                        88.5k74394






















                            up vote
                            0
                            down vote













                            To elaborate on why the technique you used in general doesn't work, it is because you introduce a new degree of freedom.



                            More precisely, given two equations
                            $$f(x,y,z)=0,$$ and $$g(x,y,z)=0,$$ if you consider the equation $$f(x,y,z)=g(x,y,z),$$ the solutions to this new equation should contain the entire intersection of the two original surfaces, but it will also contain the imtersection of the surfaces $$f(x,y,z)=1,$$ and $$g(x,y,z)=1.$$ Or, in other words, the equation $$f(x,y,z)=g(x,y,z)$$ can have solutions where $$f(x,y,z)neq 0,$$ but these solutions violate the original equations.






                            share|cite|improve this answer

























                              up vote
                              0
                              down vote













                              To elaborate on why the technique you used in general doesn't work, it is because you introduce a new degree of freedom.



                              More precisely, given two equations
                              $$f(x,y,z)=0,$$ and $$g(x,y,z)=0,$$ if you consider the equation $$f(x,y,z)=g(x,y,z),$$ the solutions to this new equation should contain the entire intersection of the two original surfaces, but it will also contain the imtersection of the surfaces $$f(x,y,z)=1,$$ and $$g(x,y,z)=1.$$ Or, in other words, the equation $$f(x,y,z)=g(x,y,z)$$ can have solutions where $$f(x,y,z)neq 0,$$ but these solutions violate the original equations.






                              share|cite|improve this answer























                                up vote
                                0
                                down vote










                                up vote
                                0
                                down vote









                                To elaborate on why the technique you used in general doesn't work, it is because you introduce a new degree of freedom.



                                More precisely, given two equations
                                $$f(x,y,z)=0,$$ and $$g(x,y,z)=0,$$ if you consider the equation $$f(x,y,z)=g(x,y,z),$$ the solutions to this new equation should contain the entire intersection of the two original surfaces, but it will also contain the imtersection of the surfaces $$f(x,y,z)=1,$$ and $$g(x,y,z)=1.$$ Or, in other words, the equation $$f(x,y,z)=g(x,y,z)$$ can have solutions where $$f(x,y,z)neq 0,$$ but these solutions violate the original equations.






                                share|cite|improve this answer












                                To elaborate on why the technique you used in general doesn't work, it is because you introduce a new degree of freedom.



                                More precisely, given two equations
                                $$f(x,y,z)=0,$$ and $$g(x,y,z)=0,$$ if you consider the equation $$f(x,y,z)=g(x,y,z),$$ the solutions to this new equation should contain the entire intersection of the two original surfaces, but it will also contain the imtersection of the surfaces $$f(x,y,z)=1,$$ and $$g(x,y,z)=1.$$ Or, in other words, the equation $$f(x,y,z)=g(x,y,z)$$ can have solutions where $$f(x,y,z)neq 0,$$ but these solutions violate the original equations.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Nov 23 at 17:03









                                Sean English

                                3,224719




                                3,224719






















                                    up vote
                                    0
                                    down vote













                                    I think that you have been badly misled by sloppy teaching at some level, probably high-school. You should never expect that you can find the common solution(s) of two equations by setting them equal, except in some rather special situations. Let me go into some detail:



                                    One case where your strategy will work is when you have two equations, $y=f(x)$ and $y=g(x)$, and want to know where the two graphs intersect. Then you may (and should) write $f(x)=g(x)$ and solve the single equation $f(x)-g(x)=0$ for $x$. The reason this works is that by setting $f=g$, you have in effect eliminated the letter $y$ from the picture, and can now use one-variable methods to come to the desired answer.



                                    Your particular problem presented you with a three-variable situation, and, as @B.Goddard and @gimusi have said in their answers, the simple methods applicable to two-variable problems have to be handled rather differently.



                                    But let me come back to the two-variable case, where you expect finitely many common points to two planar curves. There, setting $f=g$ in the case that you were intersecting two graphs was just a neat and easy trick for eliminating $y$ from the situation, and in the two-variable situation, you should always base your strategy on eliminating one of the variables. Take an interesting case like intersecting a parabola with a circle, like $y=x^2+x-2$ and $x^2+y^2=4$. Here you easily eliminate $y$ by substituting its expression in the first equation for the appearance(s) of $y$ in the second, to get
                                    begin{align}
                                    x^2+(x^2+x-2)^2&=4\
                                    x^2+x^4+2x^3-3x^2-4x+4&=4\
                                    x^4+2x^3-2x^2-4x&=0\
                                    x(x+2)(x^2-2)&=0,,
                                    end{align}

                                    to give the four(*) points $(-2,0)$, $(-sqrt2,-sqrt2)$, $(0,-2)$, and $(-sqrt2,-sqrt2)$ of the intersection.



                                    In cases where neither equation is already solved for
                                    $x$ nor $y$, you may have to try to solve one by your own means. If the equations are of higher degree than $2$, this can get tricky.



                                    (*) According to a wonderful theorem of Bézout, any two different conic sections will have exactly four points in their intersection, as long as you count properly.






                                    share|cite|improve this answer

























                                      up vote
                                      0
                                      down vote













                                      I think that you have been badly misled by sloppy teaching at some level, probably high-school. You should never expect that you can find the common solution(s) of two equations by setting them equal, except in some rather special situations. Let me go into some detail:



                                      One case where your strategy will work is when you have two equations, $y=f(x)$ and $y=g(x)$, and want to know where the two graphs intersect. Then you may (and should) write $f(x)=g(x)$ and solve the single equation $f(x)-g(x)=0$ for $x$. The reason this works is that by setting $f=g$, you have in effect eliminated the letter $y$ from the picture, and can now use one-variable methods to come to the desired answer.



                                      Your particular problem presented you with a three-variable situation, and, as @B.Goddard and @gimusi have said in their answers, the simple methods applicable to two-variable problems have to be handled rather differently.



                                      But let me come back to the two-variable case, where you expect finitely many common points to two planar curves. There, setting $f=g$ in the case that you were intersecting two graphs was just a neat and easy trick for eliminating $y$ from the situation, and in the two-variable situation, you should always base your strategy on eliminating one of the variables. Take an interesting case like intersecting a parabola with a circle, like $y=x^2+x-2$ and $x^2+y^2=4$. Here you easily eliminate $y$ by substituting its expression in the first equation for the appearance(s) of $y$ in the second, to get
                                      begin{align}
                                      x^2+(x^2+x-2)^2&=4\
                                      x^2+x^4+2x^3-3x^2-4x+4&=4\
                                      x^4+2x^3-2x^2-4x&=0\
                                      x(x+2)(x^2-2)&=0,,
                                      end{align}

                                      to give the four(*) points $(-2,0)$, $(-sqrt2,-sqrt2)$, $(0,-2)$, and $(-sqrt2,-sqrt2)$ of the intersection.



                                      In cases where neither equation is already solved for
                                      $x$ nor $y$, you may have to try to solve one by your own means. If the equations are of higher degree than $2$, this can get tricky.



                                      (*) According to a wonderful theorem of Bézout, any two different conic sections will have exactly four points in their intersection, as long as you count properly.






                                      share|cite|improve this answer























                                        up vote
                                        0
                                        down vote










                                        up vote
                                        0
                                        down vote









                                        I think that you have been badly misled by sloppy teaching at some level, probably high-school. You should never expect that you can find the common solution(s) of two equations by setting them equal, except in some rather special situations. Let me go into some detail:



                                        One case where your strategy will work is when you have two equations, $y=f(x)$ and $y=g(x)$, and want to know where the two graphs intersect. Then you may (and should) write $f(x)=g(x)$ and solve the single equation $f(x)-g(x)=0$ for $x$. The reason this works is that by setting $f=g$, you have in effect eliminated the letter $y$ from the picture, and can now use one-variable methods to come to the desired answer.



                                        Your particular problem presented you with a three-variable situation, and, as @B.Goddard and @gimusi have said in their answers, the simple methods applicable to two-variable problems have to be handled rather differently.



                                        But let me come back to the two-variable case, where you expect finitely many common points to two planar curves. There, setting $f=g$ in the case that you were intersecting two graphs was just a neat and easy trick for eliminating $y$ from the situation, and in the two-variable situation, you should always base your strategy on eliminating one of the variables. Take an interesting case like intersecting a parabola with a circle, like $y=x^2+x-2$ and $x^2+y^2=4$. Here you easily eliminate $y$ by substituting its expression in the first equation for the appearance(s) of $y$ in the second, to get
                                        begin{align}
                                        x^2+(x^2+x-2)^2&=4\
                                        x^2+x^4+2x^3-3x^2-4x+4&=4\
                                        x^4+2x^3-2x^2-4x&=0\
                                        x(x+2)(x^2-2)&=0,,
                                        end{align}

                                        to give the four(*) points $(-2,0)$, $(-sqrt2,-sqrt2)$, $(0,-2)$, and $(-sqrt2,-sqrt2)$ of the intersection.



                                        In cases where neither equation is already solved for
                                        $x$ nor $y$, you may have to try to solve one by your own means. If the equations are of higher degree than $2$, this can get tricky.



                                        (*) According to a wonderful theorem of Bézout, any two different conic sections will have exactly four points in their intersection, as long as you count properly.






                                        share|cite|improve this answer












                                        I think that you have been badly misled by sloppy teaching at some level, probably high-school. You should never expect that you can find the common solution(s) of two equations by setting them equal, except in some rather special situations. Let me go into some detail:



                                        One case where your strategy will work is when you have two equations, $y=f(x)$ and $y=g(x)$, and want to know where the two graphs intersect. Then you may (and should) write $f(x)=g(x)$ and solve the single equation $f(x)-g(x)=0$ for $x$. The reason this works is that by setting $f=g$, you have in effect eliminated the letter $y$ from the picture, and can now use one-variable methods to come to the desired answer.



                                        Your particular problem presented you with a three-variable situation, and, as @B.Goddard and @gimusi have said in their answers, the simple methods applicable to two-variable problems have to be handled rather differently.



                                        But let me come back to the two-variable case, where you expect finitely many common points to two planar curves. There, setting $f=g$ in the case that you were intersecting two graphs was just a neat and easy trick for eliminating $y$ from the situation, and in the two-variable situation, you should always base your strategy on eliminating one of the variables. Take an interesting case like intersecting a parabola with a circle, like $y=x^2+x-2$ and $x^2+y^2=4$. Here you easily eliminate $y$ by substituting its expression in the first equation for the appearance(s) of $y$ in the second, to get
                                        begin{align}
                                        x^2+(x^2+x-2)^2&=4\
                                        x^2+x^4+2x^3-3x^2-4x+4&=4\
                                        x^4+2x^3-2x^2-4x&=0\
                                        x(x+2)(x^2-2)&=0,,
                                        end{align}

                                        to give the four(*) points $(-2,0)$, $(-sqrt2,-sqrt2)$, $(0,-2)$, and $(-sqrt2,-sqrt2)$ of the intersection.



                                        In cases where neither equation is already solved for
                                        $x$ nor $y$, you may have to try to solve one by your own means. If the equations are of higher degree than $2$, this can get tricky.



                                        (*) According to a wonderful theorem of Bézout, any two different conic sections will have exactly four points in their intersection, as long as you count properly.







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                                        answered Nov 23 at 18:10









                                        Lubin

                                        43.1k44385




                                        43.1k44385






















                                            up vote
                                            0
                                            down vote













                                            Eliminating $x$ we get $(y-z)$ orthogonal plane projection ( and eliminating $y$ gives $(x-y)$ plane projection) etc.



                                            A parameter can be chosen to represent the three coordinates of the curve of intersection by eliminating or isolating one variable.



                                            But by directly adding the two equations together in linear combinations as in an example



                                            $$ lambda(x^2+y^2) + (lambda-1) (x - y + z +2) =0 $$



                                            we get a set or family of second degree conicoids/quadrics passing through their common intersection curve ( ellipse in this case). For equal weightage as in the case $lambda=frac12$ you mention, it is a paraboloid.






                                            share|cite|improve this answer



























                                              up vote
                                              0
                                              down vote













                                              Eliminating $x$ we get $(y-z)$ orthogonal plane projection ( and eliminating $y$ gives $(x-y)$ plane projection) etc.



                                              A parameter can be chosen to represent the three coordinates of the curve of intersection by eliminating or isolating one variable.



                                              But by directly adding the two equations together in linear combinations as in an example



                                              $$ lambda(x^2+y^2) + (lambda-1) (x - y + z +2) =0 $$



                                              we get a set or family of second degree conicoids/quadrics passing through their common intersection curve ( ellipse in this case). For equal weightage as in the case $lambda=frac12$ you mention, it is a paraboloid.






                                              share|cite|improve this answer

























                                                up vote
                                                0
                                                down vote










                                                up vote
                                                0
                                                down vote









                                                Eliminating $x$ we get $(y-z)$ orthogonal plane projection ( and eliminating $y$ gives $(x-y)$ plane projection) etc.



                                                A parameter can be chosen to represent the three coordinates of the curve of intersection by eliminating or isolating one variable.



                                                But by directly adding the two equations together in linear combinations as in an example



                                                $$ lambda(x^2+y^2) + (lambda-1) (x - y + z +2) =0 $$



                                                we get a set or family of second degree conicoids/quadrics passing through their common intersection curve ( ellipse in this case). For equal weightage as in the case $lambda=frac12$ you mention, it is a paraboloid.






                                                share|cite|improve this answer














                                                Eliminating $x$ we get $(y-z)$ orthogonal plane projection ( and eliminating $y$ gives $(x-y)$ plane projection) etc.



                                                A parameter can be chosen to represent the three coordinates of the curve of intersection by eliminating or isolating one variable.



                                                But by directly adding the two equations together in linear combinations as in an example



                                                $$ lambda(x^2+y^2) + (lambda-1) (x - y + z +2) =0 $$



                                                we get a set or family of second degree conicoids/quadrics passing through their common intersection curve ( ellipse in this case). For equal weightage as in the case $lambda=frac12$ you mention, it is a paraboloid.







                                                share|cite|improve this answer














                                                share|cite|improve this answer



                                                share|cite|improve this answer








                                                edited Nov 23 at 18:22

























                                                answered Nov 23 at 18:11









                                                Narasimham

                                                20.4k52158




                                                20.4k52158






























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