Parametrization of the intersection of a cone and plane.
Clash Royale CLAN TAG#URR8PPP
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EDITED with new progress updates.
As the title states, I'm trying to parametrize the intersection of a cone and a plane. The equations are:
$z^2 = 2x^2+2y^2$ and
$2x+y+3z=4implies z=frac13(4-2x-y)$
If it were either a plane parallel to the x-y plane or a cylinder instead of a cone, I would simply project the intersection onto the x-y plane, parametrize the projection for x and y then insert those equations into the equation of the plane to get the parametrization of z. I'm thinking a similar approach would work here but I can't figure out how to find the equation of the projection.
I now have what I believe is the projected ellipse onto the x-y plane. The ellipse equation is:
$0 = frac19(4-2x-y)^2-2x^2-2y^2$
My problem is now that if I expand out the squared portion, I end up with $4xy$ in the equation. I have no idea how to parametrize an ellipse with a cross term in it. Could someone help with that?
Any help would be greatly appreciated.
Thank you,
Eric
multivariable-calculus parametric
asked Jan 12 '15 at 5:00
John
403211
add a comment |Â
up vote
1
down vote
favorite
EDITED with new progress updates.
As the title states, I'm trying to parametrize the intersection of a cone and a plane. The equations are:
$z^2 = 2x^2+2y^2$ and
$2x+y+3z=4implies z=frac13(4-2x-y)$
If it were either a plane parallel to the x-y plane or a cylinder instead of a cone, I would simply project the intersection onto the x-y plane, parametrize the projection for x and y then insert those equations into the equation of the plane to get the parametrization of z. I'm thinking a similar approach would work here but I can't figure out how to find the equation of the projection.
I now have what I believe is the projected ellipse onto the x-y plane. The ellipse equation is:
$0 = frac19(4-2x-y)^2-2x^2-2y^2$
My problem is now that if I expand out the squared portion, I end up with $4xy$ in the equation. I have no idea how to parametrize an ellipse with a cross term in it. Could someone help with that?
Any help would be greatly appreciated.
Thank you,
Eric
multivariable-calculus parametric
asked Jan 12 '15 at 5:00
John
403211
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
EDITED with new progress updates.
As the title states, I'm trying to parametrize the intersection of a cone and a plane. The equations are:
$z^2 = 2x^2+2y^2$ and
$2x+y+3z=4implies z=frac13(4-2x-y)$
If it were either a plane parallel to the x-y plane or a cylinder instead of a cone, I would simply project the intersection onto the x-y plane, parametrize the projection for x and y then insert those equations into the equation of the plane to get the parametrization of z. I'm thinking a similar approach would work here but I can't figure out how to find the equation of the projection.
I now have what I believe is the projected ellipse onto the x-y plane. The ellipse equation is:
$0 = frac19(4-2x-y)^2-2x^2-2y^2$
My problem is now that if I expand out the squared portion, I end up with $4xy$ in the equation. I have no idea how to parametrize an ellipse with a cross term in it. Could someone help with that?
Any help would be greatly appreciated.
Thank you,
Eric
multivariable-calculus parametric
asked Jan 12 '15 at 5:00
John
403211
EDITED with new progress updates.
As the title states, I'm trying to parametrize the intersection of a cone and a plane. The equations are:
$z^2 = 2x^2+2y^2$ and
$2x+y+3z=4implies z=frac13(4-2x-y)$
If it were either a plane parallel to the x-y plane or a cylinder instead of a cone, I would simply project the intersection onto the x-y plane, parametrize the projection for x and y then insert those equations into the equation of the plane to get the parametrization of z. I'm thinking a similar approach would work here but I can't figure out how to find the equation of the projection.
I now have what I believe is the projected ellipse onto the x-y plane. The ellipse equation is:
$0 = frac19(4-2x-y)^2-2x^2-2y^2$
My problem is now that if I expand out the squared portion, I end up with $4xy$ in the equation. I have no idea how to parametrize an ellipse with a cross term in it. Could someone help with that?
Any help would be greatly appreciated.
Thank you,
Eric
multivariable-calculus parametric
asked Jan 12 '15 at 5:00
John
403211
edited Jan 12 '15 at 5:59
asked Jan 12 '15 at 5:00
John
403211
asked Jan 12 '15 at 5:00
John
403211
asked Jan 12 '15 at 5:00
John
403211
403211
add a comment |Â
add a comment |Â
2 Answers
2
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0
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Hint: eliminate one of the variables by solving the plane equation for it and substituting into the cone equation.
EDIT: For an ellipse equation with "cross terms", one thing you can do is "complete the square" in either the $x$ or $y$, so it becomes something like
$a (x + b y)^2 + c y^2 = 1$ with $a, c > 0$. Then you can parametrize it as
$$eqaligny &= dfracsin(t)sqrtccr x &= -b y + dfraccos(t)sqrta
= - dfracb sin(t)sqrtc + dfraccos(t)sqrtacr$$
answered Jan 12 '15 at 5:06
Robert Israel
305k22201443
Hello Robert. Thank you, I've actually been trying that but, unless I've been doing something wrong, I keep ending up with cross terms, e.g. $4xy$ and I'm not sure how to handle those without using a rotation matrix of some sort but this problem doesn't have any linear algebra knowledge requirements. Should I not be getting any cross terms?
– John
Jan 12 '15 at 5:11
When you expand it out you get an ellipse but it's rotated which is why there are xy terms. I can post an answer with the equation if you'd like.
– Gabriel
Jan 12 '15 at 5:13
@Gabriel Ok, so I stopped worrying about the cross term and got the equation for the ellipse that looks like it is a good projection. Unfortunately, I'm now stuck trying to parametrize this ellipse equation that has cross terms.
– John
Jan 12 '15 at 5:30
add a comment |Â
up vote
0
down vote
As @RobertIsrael suggests, take $z$ as a parameter and solve
$$2x^2+2y^2=z^2,\2x+y=4-3z.$$
The second equation gives
$$y=4-2x-3z,$$ and plugging in the first
$$10x^2+24xz-32x+18z^2-48z+32=z^2,$$ giving the solutions in $x$ and $y$
$$x=frac85pmfracsqrt-26z^2+96z-6410-frac6z5,\
y=frac45mpfracsqrt-26z^2+96z-645-frac3z5.$$
Seeing the polynomial under the radical, this is an ellipse. If you don't like the double signs, you can complete the square
$$-26z^2+96z-64=26left(frac94413-left(z-frac4813right)^2right)$$ and set
$$z=sqrtfrac94413cos t+frac4813.$$
This will result in $x,y,z$ being all three affine combinations of $cos t,sin t$.
answered Jul 29 '16 at 8:44
Yves Daoust
112k665205
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Hint: eliminate one of the variables by solving the plane equation for it and substituting into the cone equation.
EDIT: For an ellipse equation with "cross terms", one thing you can do is "complete the square" in either the $x$ or $y$, so it becomes something like
$a (x + b y)^2 + c y^2 = 1$ with $a, c > 0$. Then you can parametrize it as
$$eqaligny &= dfracsin(t)sqrtccr x &= -b y + dfraccos(t)sqrta
= - dfracb sin(t)sqrtc + dfraccos(t)sqrtacr$$
answered Jan 12 '15 at 5:06
Robert Israel
305k22201443
Hello Robert. Thank you, I've actually been trying that but, unless I've been doing something wrong, I keep ending up with cross terms, e.g. $4xy$ and I'm not sure how to handle those without using a rotation matrix of some sort but this problem doesn't have any linear algebra knowledge requirements. Should I not be getting any cross terms?
– John
Jan 12 '15 at 5:11
When you expand it out you get an ellipse but it's rotated which is why there are xy terms. I can post an answer with the equation if you'd like.
– Gabriel
Jan 12 '15 at 5:13
@Gabriel Ok, so I stopped worrying about the cross term and got the equation for the ellipse that looks like it is a good projection. Unfortunately, I'm now stuck trying to parametrize this ellipse equation that has cross terms.
– John
Jan 12 '15 at 5:30
add a comment |Â
up vote
0
down vote
Hint: eliminate one of the variables by solving the plane equation for it and substituting into the cone equation.
EDIT: For an ellipse equation with "cross terms", one thing you can do is "complete the square" in either the $x$ or $y$, so it becomes something like
$a (x + b y)^2 + c y^2 = 1$ with $a, c > 0$. Then you can parametrize it as
$$eqaligny &= dfracsin(t)sqrtccr x &= -b y + dfraccos(t)sqrta
= - dfracb sin(t)sqrtc + dfraccos(t)sqrtacr$$
answered Jan 12 '15 at 5:06
Robert Israel
305k22201443
Hello Robert. Thank you, I've actually been trying that but, unless I've been doing something wrong, I keep ending up with cross terms, e.g. $4xy$ and I'm not sure how to handle those without using a rotation matrix of some sort but this problem doesn't have any linear algebra knowledge requirements. Should I not be getting any cross terms?
– John
Jan 12 '15 at 5:11
When you expand it out you get an ellipse but it's rotated which is why there are xy terms. I can post an answer with the equation if you'd like.
– Gabriel
Jan 12 '15 at 5:13
@Gabriel Ok, so I stopped worrying about the cross term and got the equation for the ellipse that looks like it is a good projection. Unfortunately, I'm now stuck trying to parametrize this ellipse equation that has cross terms.
– John
Jan 12 '15 at 5:30
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: eliminate one of the variables by solving the plane equation for it and substituting into the cone equation.
EDIT: For an ellipse equation with "cross terms", one thing you can do is "complete the square" in either the $x$ or $y$, so it becomes something like
$a (x + b y)^2 + c y^2 = 1$ with $a, c > 0$. Then you can parametrize it as
$$eqaligny &= dfracsin(t)sqrtccr x &= -b y + dfraccos(t)sqrta
= - dfracb sin(t)sqrtc + dfraccos(t)sqrtacr$$
answered Jan 12 '15 at 5:06
Robert Israel
305k22201443
Hint: eliminate one of the variables by solving the plane equation for it and substituting into the cone equation.
EDIT: For an ellipse equation with "cross terms", one thing you can do is "complete the square" in either the $x$ or $y$, so it becomes something like
$a (x + b y)^2 + c y^2 = 1$ with $a, c > 0$. Then you can parametrize it as
$$eqaligny &= dfracsin(t)sqrtccr x &= -b y + dfraccos(t)sqrta
= - dfracb sin(t)sqrtc + dfraccos(t)sqrtacr$$
answered Jan 12 '15 at 5:06
Robert Israel
305k22201443
edited Jan 12 '15 at 7:05
answered Jan 12 '15 at 5:06
Robert Israel
305k22201443
answered Jan 12 '15 at 5:06
Robert Israel
305k22201443
answered Jan 12 '15 at 5:06
Robert Israel
305k22201443
305k22201443
Hello Robert. Thank you, I've actually been trying that but, unless I've been doing something wrong, I keep ending up with cross terms, e.g. $4xy$ and I'm not sure how to handle those without using a rotation matrix of some sort but this problem doesn't have any linear algebra knowledge requirements. Should I not be getting any cross terms?
– John
Jan 12 '15 at 5:11
When you expand it out you get an ellipse but it's rotated which is why there are xy terms. I can post an answer with the equation if you'd like.
– Gabriel
Jan 12 '15 at 5:13
@Gabriel Ok, so I stopped worrying about the cross term and got the equation for the ellipse that looks like it is a good projection. Unfortunately, I'm now stuck trying to parametrize this ellipse equation that has cross terms.
– John
Jan 12 '15 at 5:30
add a comment |Â
Hello Robert. Thank you, I've actually been trying that but, unless I've been doing something wrong, I keep ending up with cross terms, e.g. $4xy$ and I'm not sure how to handle those without using a rotation matrix of some sort but this problem doesn't have any linear algebra knowledge requirements. Should I not be getting any cross terms?
– John
Jan 12 '15 at 5:11
When you expand it out you get an ellipse but it's rotated which is why there are xy terms. I can post an answer with the equation if you'd like.
– Gabriel
Jan 12 '15 at 5:13
@Gabriel Ok, so I stopped worrying about the cross term and got the equation for the ellipse that looks like it is a good projection. Unfortunately, I'm now stuck trying to parametrize this ellipse equation that has cross terms.
– John
Jan 12 '15 at 5:30
Hello Robert. Thank you, I've actually been trying that but, unless I've been doing something wrong, I keep ending up with cross terms, e.g. $4xy$ and I'm not sure how to handle those without using a rotation matrix of some sort but this problem doesn't have any linear algebra knowledge requirements. Should I not be getting any cross terms?
– John
Jan 12 '15 at 5:11
Hello Robert. Thank you, I've actually been trying that but, unless I've been doing something wrong, I keep ending up with cross terms, e.g. $4xy$ and I'm not sure how to handle those without using a rotation matrix of some sort but this problem doesn't have any linear algebra knowledge requirements. Should I not be getting any cross terms?
– John
Jan 12 '15 at 5:11
When you expand it out you get an ellipse but it's rotated which is why there are xy terms. I can post an answer with the equation if you'd like.
– Gabriel
Jan 12 '15 at 5:13
When you expand it out you get an ellipse but it's rotated which is why there are xy terms. I can post an answer with the equation if you'd like.
– Gabriel
Jan 12 '15 at 5:13
@Gabriel Ok, so I stopped worrying about the cross term and got the equation for the ellipse that looks like it is a good projection. Unfortunately, I'm now stuck trying to parametrize this ellipse equation that has cross terms.
– John
Jan 12 '15 at 5:30
@Gabriel Ok, so I stopped worrying about the cross term and got the equation for the ellipse that looks like it is a good projection. Unfortunately, I'm now stuck trying to parametrize this ellipse equation that has cross terms.
– John
Jan 12 '15 at 5:30
add a comment |Â
up vote
0
down vote
As @RobertIsrael suggests, take $z$ as a parameter and solve
$$2x^2+2y^2=z^2,\2x+y=4-3z.$$
The second equation gives
$$y=4-2x-3z,$$ and plugging in the first
$$10x^2+24xz-32x+18z^2-48z+32=z^2,$$ giving the solutions in $x$ and $y$
$$x=frac85pmfracsqrt-26z^2+96z-6410-frac6z5,\
y=frac45mpfracsqrt-26z^2+96z-645-frac3z5.$$
Seeing the polynomial under the radical, this is an ellipse. If you don't like the double signs, you can complete the square
$$-26z^2+96z-64=26left(frac94413-left(z-frac4813right)^2right)$$ and set
$$z=sqrtfrac94413cos t+frac4813.$$
This will result in $x,y,z$ being all three affine combinations of $cos t,sin t$.
answered Jul 29 '16 at 8:44
Yves Daoust
112k665205
add a comment |Â
up vote
0
down vote
As @RobertIsrael suggests, take $z$ as a parameter and solve
$$2x^2+2y^2=z^2,\2x+y=4-3z.$$
The second equation gives
$$y=4-2x-3z,$$ and plugging in the first
$$10x^2+24xz-32x+18z^2-48z+32=z^2,$$ giving the solutions in $x$ and $y$
$$x=frac85pmfracsqrt-26z^2+96z-6410-frac6z5,\
y=frac45mpfracsqrt-26z^2+96z-645-frac3z5.$$
Seeing the polynomial under the radical, this is an ellipse. If you don't like the double signs, you can complete the square
$$-26z^2+96z-64=26left(frac94413-left(z-frac4813right)^2right)$$ and set
$$z=sqrtfrac94413cos t+frac4813.$$
This will result in $x,y,z$ being all three affine combinations of $cos t,sin t$.
answered Jul 29 '16 at 8:44
Yves Daoust
112k665205
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As @RobertIsrael suggests, take $z$ as a parameter and solve
$$2x^2+2y^2=z^2,\2x+y=4-3z.$$
The second equation gives
$$y=4-2x-3z,$$ and plugging in the first
$$10x^2+24xz-32x+18z^2-48z+32=z^2,$$ giving the solutions in $x$ and $y$
$$x=frac85pmfracsqrt-26z^2+96z-6410-frac6z5,\
y=frac45mpfracsqrt-26z^2+96z-645-frac3z5.$$
Seeing the polynomial under the radical, this is an ellipse. If you don't like the double signs, you can complete the square
$$-26z^2+96z-64=26left(frac94413-left(z-frac4813right)^2right)$$ and set
$$z=sqrtfrac94413cos t+frac4813.$$
This will result in $x,y,z$ being all three affine combinations of $cos t,sin t$.
answered Jul 29 '16 at 8:44
Yves Daoust
112k665205
As @RobertIsrael suggests, take $z$ as a parameter and solve
$$2x^2+2y^2=z^2,\2x+y=4-3z.$$
The second equation gives
$$y=4-2x-3z,$$ and plugging in the first
$$10x^2+24xz-32x+18z^2-48z+32=z^2,$$ giving the solutions in $x$ and $y$
$$x=frac85pmfracsqrt-26z^2+96z-6410-frac6z5,\
y=frac45mpfracsqrt-26z^2+96z-645-frac3z5.$$
Seeing the polynomial under the radical, this is an ellipse. If you don't like the double signs, you can complete the square
$$-26z^2+96z-64=26left(frac94413-left(z-frac4813right)^2right)$$ and set
$$z=sqrtfrac94413cos t+frac4813.$$
This will result in $x,y,z$ being all three affine combinations of $cos t,sin t$.
answered Jul 29 '16 at 8:44
Yves Daoust
112k665205
answered Jul 29 '16 at 8:44
Yves Daoust
112k665205
answered Jul 29 '16 at 8:44
Yves Daoust
112k665205
answered Jul 29 '16 at 8:44
Yves Daoust
112k665205
112k665205
add a comment |Â
add a comment |Â
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