Mixing
How to find the equation of the conic before applying the rotation?
$begingroup$
Given the rotation matrix:
$$Q=begin{bmatrix}frac{2}{sqrt{5}}& frac{1}{sqrt{5}}\-frac{1}{sqrt{5}}& frac{2}{sqrt{5}}end{bmatrix},$$ I want to find the equation of the conic $C$ given by $$x^2+4xy+4y^2-10sqrt{5}x=0,$$ which I know is a parabola, before applying the rotation to it.
I was thinking about solving the system of equation $$begin{cases}x=frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}\ y=-frac{1}{sqrt{5}}tilde{x}+frac{2}{sqrt{5}}tilde{y}end{cases},$$ and then substituting $x$ and $y$ in the initial equation of the conic. The result should be $$5tilde{y}^2-20tilde{x}-10tilde{y}=0,$$ but it seems that my results are different.
linear-algebra matrices conic-sections rotations
edited Jan 21 at 22:00
jordan_glen
1
asked Jan 21 at 20:28
KevinKevin
16211
$endgroup$
|
show 5 more comments
$begingroup$
Given the rotation matrix:
$$Q=begin{bmatrix}frac{2}{sqrt{5}}& frac{1}{sqrt{5}}\-frac{1}{sqrt{5}}& frac{2}{sqrt{5}}end{bmatrix},$$ I want to find the equation of the conic $C$ given by $$x^2+4xy+4y^2-10sqrt{5}x=0,$$ which I know is a parabola, before applying the rotation to it.
I was thinking about solving the system of equation $$begin{cases}x=frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}\ y=-frac{1}{sqrt{5}}tilde{x}+frac{2}{sqrt{5}}tilde{y}end{cases},$$ and then substituting $x$ and $y$ in the initial equation of the conic. The result should be $$5tilde{y}^2-20tilde{x}-10tilde{y}=0,$$ but it seems that my results are different.
linear-algebra matrices conic-sections rotations
edited Jan 21 at 22:00
jordan_glen
1
asked Jan 21 at 20:28
KevinKevin
16211
$endgroup$
$begingroup$
You mean you want to find the equation after applying the rotation? Before applying it, you have the equation that you wrote down, or am I misunderstanding?
$endgroup$
– Servaes
Jan 21 at 20:48
1
$begingroup$
I'm not sure but you may be applying the inverse rotation.
$endgroup$
– Harnak
Jan 21 at 20:49
$begingroup$
Another way, substantially equivalent but slighty easier in calculations, is to consider $(x,y)^tA(x,y)+Bcdot(x,y)$, and then applying $(tilde{x},tilde{y})=Q(x,y)$
$endgroup$
– gabriele cassese
Jan 21 at 20:49
$begingroup$
@Servaes No, the equation of the conic has been rotated by the rotation matrix. Now, I want to find the equation of the conic before the rotation.
$endgroup$
– Kevin
Jan 21 at 20:50
$begingroup$
@amd $-10sqrt{5}x$
$endgroup$
– gabriele cassese
Jan 21 at 20:51
|
show 5 more comments
$begingroup$
Given the rotation matrix:
$$Q=begin{bmatrix}frac{2}{sqrt{5}}& frac{1}{sqrt{5}}\-frac{1}{sqrt{5}}& frac{2}{sqrt{5}}end{bmatrix},$$ I want to find the equation of the conic $C$ given by $$x^2+4xy+4y^2-10sqrt{5}x=0,$$ which I know is a parabola, before applying the rotation to it.
I was thinking about solving the system of equation $$begin{cases}x=frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}\ y=-frac{1}{sqrt{5}}tilde{x}+frac{2}{sqrt{5}}tilde{y}end{cases},$$ and then substituting $x$ and $y$ in the initial equation of the conic. The result should be $$5tilde{y}^2-20tilde{x}-10tilde{y}=0,$$ but it seems that my results are different.
linear-algebra matrices conic-sections rotations
edited Jan 21 at 22:00
jordan_glen
1
asked Jan 21 at 20:28
KevinKevin
16211
$endgroup$
Given the rotation matrix:
$$Q=begin{bmatrix}frac{2}{sqrt{5}}& frac{1}{sqrt{5}}\-frac{1}{sqrt{5}}& frac{2}{sqrt{5}}end{bmatrix},$$ I want to find the equation of the conic $C$ given by $$x^2+4xy+4y^2-10sqrt{5}x=0,$$ which I know is a parabola, before applying the rotation to it.
I was thinking about solving the system of equation $$begin{cases}x=frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}\ y=-frac{1}{sqrt{5}}tilde{x}+frac{2}{sqrt{5}}tilde{y}end{cases},$$ and then substituting $x$ and $y$ in the initial equation of the conic. The result should be $$5tilde{y}^2-20tilde{x}-10tilde{y}=0,$$ but it seems that my results are different.
linear-algebra matrices conic-sections rotations
linear-algebra matrices conic-sections rotations
edited Jan 21 at 22:00
jordan_glen
1
asked Jan 21 at 20:28
KevinKevin
16211
edited Jan 21 at 22:00
jordan_glen
1
asked Jan 21 at 20:28
KevinKevin
16211
edited Jan 21 at 22:00
jordan_glen
1
edited Jan 21 at 22:00
jordan_glen
1
edited Jan 21 at 22:00
jordan_glen
1
1
asked Jan 21 at 20:28
KevinKevin
16211
asked Jan 21 at 20:28
KevinKevin
16211
asked Jan 21 at 20:28
KevinKevin
16211
16211
$begingroup$
You mean you want to find the equation after applying the rotation? Before applying it, you have the equation that you wrote down, or am I misunderstanding?
$endgroup$
– Servaes
Jan 21 at 20:48
1
$begingroup$
I'm not sure but you may be applying the inverse rotation.
$endgroup$
– Harnak
Jan 21 at 20:49
$begingroup$
Another way, substantially equivalent but slighty easier in calculations, is to consider $(x,y)^tA(x,y)+Bcdot(x,y)$, and then applying $(tilde{x},tilde{y})=Q(x,y)$
$endgroup$
– gabriele cassese
Jan 21 at 20:49
$begingroup$
@Servaes No, the equation of the conic has been rotated by the rotation matrix. Now, I want to find the equation of the conic before the rotation.
$endgroup$
– Kevin
Jan 21 at 20:50
$begingroup$
@amd $-10sqrt{5}x$
$endgroup$
– gabriele cassese
Jan 21 at 20:51
|
show 5 more comments
$begingroup$
You mean you want to find the equation after applying the rotation? Before applying it, you have the equation that you wrote down, or am I misunderstanding?
$endgroup$
– Servaes
Jan 21 at 20:48
1
$begingroup$
I'm not sure but you may be applying the inverse rotation.
$endgroup$
– Harnak
Jan 21 at 20:49
$begingroup$
Another way, substantially equivalent but slighty easier in calculations, is to consider $(x,y)^tA(x,y)+Bcdot(x,y)$, and then applying $(tilde{x},tilde{y})=Q(x,y)$
$endgroup$
– gabriele cassese
Jan 21 at 20:49
$begingroup$
@Servaes No, the equation of the conic has been rotated by the rotation matrix. Now, I want to find the equation of the conic before the rotation.
$endgroup$
– Kevin
Jan 21 at 20:50
$begingroup$
@amd $-10sqrt{5}x$
$endgroup$
– gabriele cassese
Jan 21 at 20:51
$begingroup$
You mean you want to find the equation after applying the rotation? Before applying it, you have the equation that you wrote down, or am I misunderstanding?
$endgroup$
– Servaes
Jan 21 at 20:48
$begingroup$
You mean you want to find the equation after applying the rotation? Before applying it, you have the equation that you wrote down, or am I misunderstanding?
$endgroup$
– Servaes
Jan 21 at 20:48
1
1
$begingroup$
I'm not sure but you may be applying the inverse rotation.
$endgroup$
– Harnak
Jan 21 at 20:49
$begingroup$
I'm not sure but you may be applying the inverse rotation.
$endgroup$
– Harnak
Jan 21 at 20:49
$begingroup$
Another way, substantially equivalent but slighty easier in calculations, is to consider $(x,y)^tA(x,y)+Bcdot(x,y)$, and then applying $(tilde{x},tilde{y})=Q(x,y)$
$endgroup$
– gabriele cassese
Jan 21 at 20:49
$begingroup$
Another way, substantially equivalent but slighty easier in calculations, is to consider $(x,y)^tA(x,y)+Bcdot(x,y)$, and then applying $(tilde{x},tilde{y})=Q(x,y)$
$endgroup$
– gabriele cassese
Jan 21 at 20:49
$begingroup$
@Servaes No, the equation of the conic has been rotated by the rotation matrix. Now, I want to find the equation of the conic before the rotation.
$endgroup$
– Kevin
Jan 21 at 20:50
$begingroup$
@Servaes No, the equation of the conic has been rotated by the rotation matrix. Now, I want to find the equation of the conic before the rotation.
$endgroup$
– Kevin
Jan 21 at 20:50
$begingroup$
@amd $-10sqrt{5}x$
$endgroup$
– gabriele cassese
Jan 21 at 20:51
$begingroup$
@amd $-10sqrt{5}x$
$endgroup$
– gabriele cassese
Jan 21 at 20:51
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Your approach is entirely correct; plugging in your substitutions into the first equation for $C$ yields the result you claim it should be:
begin{eqnarray*}
x^2+4xy+4y^2-10sqrt{5}x&=&
left(frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}right)^2+4left(frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}right)left(-frac{1}{sqrt{5}}tilde{x}+frac{2}{sqrt{5}}tilde{y}right)\
& &+4left(-frac{1}{sqrt{5}}tilde{x}+frac{2}{sqrt{5}}tilde{y}right)^2-10sqrt{5}left(frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}right)\
&=&frac{1}{5}(4tilde{x}^2+4tilde{x}tilde{y}+tilde{y}^2)+frac{4}{5}(-2tilde{x}^2+3tilde{x}tilde{y}+2tilde{y}^2)\
& &+frac{4}{5}(tilde{x}^2-4tilde{x}tilde{y}+4tilde{y}^2)-10(2tilde{x}+tilde{y})\
&=&5tilde{y}^2-20tilde{x}-10tilde{y}.
end{eqnarray*}
You seem to have made a mistake in your calculation.
answered Jan 21 at 20:50
ServaesServaes
27.5k34098
$endgroup$
add a comment |
$begingroup$
Basically, it looks like you’re getting the transformation backwards. If you have a coordinate transformation $mathbf x'=Qmathbf x$, to get the transformed equation you have to substitute for $mathbf x$ in the original equation, i.e., apply the inverse transformation to $x$ and $y$ in the equation. So, if the original coordinate transformation was given by $Q$, you now want to transform by $Q^{-1}$, which means that you substitute $Q(x,y)^T$ for $(x,y)$ in the equation that you have.
answered Jan 21 at 21:00
amdamd
30.8k21051
$endgroup$
add a comment |
$begingroup$
Suppose we have a conic with the equation $P(x,y) = 0.$
Next, you apply the rotation $Q$ to the conic,
so if $(tilde x,tilde y)$ is a point on the conic before rotation,
it is taken to the point
$$
(x,y)=left(tfrac{2}{sqrt5}tilde{x}+tfrac{1}{sqrt5}tilde{y},-tfrac{1}{sqrt5}tilde{x}+tfrac{2}{sqrt5}tilde{y}right) tag1
$$
after rotation.
Moreover, after applying the rotation, the rotated conic has the equation
$$ x^2+4xy+4y^2-10sqrt{5}x=0. tag2$$
If you now use Equation $(1)$ to provide the values of $x$ and $y$
in terms of $tilde x$ and $tilde y,$
you can make those substitutions for $x$ and $y$ in Equation $(2),$
and then you have an equation that must be satisfied by any point
$(tilde x,tilde y)$ on the original conic.
Try it.
The fact that one of your steps was to "solve" Equation $(1)$ suggests that your substitution was actually for a rotation by the correct angular magnitude, but in the opposite direction from what you needed.
answered Jan 21 at 21:16
David KDavid K
54.9k344120
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your approach is entirely correct; plugging in your substitutions into the first equation for $C$ yields the result you claim it should be:
begin{eqnarray*}
x^2+4xy+4y^2-10sqrt{5}x&=&
left(frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}right)^2+4left(frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}right)left(-frac{1}{sqrt{5}}tilde{x}+frac{2}{sqrt{5}}tilde{y}right)\
& &+4left(-frac{1}{sqrt{5}}tilde{x}+frac{2}{sqrt{5}}tilde{y}right)^2-10sqrt{5}left(frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}right)\
&=&frac{1}{5}(4tilde{x}^2+4tilde{x}tilde{y}+tilde{y}^2)+frac{4}{5}(-2tilde{x}^2+3tilde{x}tilde{y}+2tilde{y}^2)\
& &+frac{4}{5}(tilde{x}^2-4tilde{x}tilde{y}+4tilde{y}^2)-10(2tilde{x}+tilde{y})\
&=&5tilde{y}^2-20tilde{x}-10tilde{y}.
end{eqnarray*}
You seem to have made a mistake in your calculation.
answered Jan 21 at 20:50
ServaesServaes
27.5k34098
$endgroup$
add a comment |
$begingroup$
Your approach is entirely correct; plugging in your substitutions into the first equation for $C$ yields the result you claim it should be:
begin{eqnarray*}
x^2+4xy+4y^2-10sqrt{5}x&=&
left(frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}right)^2+4left(frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}right)left(-frac{1}{sqrt{5}}tilde{x}+frac{2}{sqrt{5}}tilde{y}right)\
& &+4left(-frac{1}{sqrt{5}}tilde{x}+frac{2}{sqrt{5}}tilde{y}right)^2-10sqrt{5}left(frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}right)\
&=&frac{1}{5}(4tilde{x}^2+4tilde{x}tilde{y}+tilde{y}^2)+frac{4}{5}(-2tilde{x}^2+3tilde{x}tilde{y}+2tilde{y}^2)\
& &+frac{4}{5}(tilde{x}^2-4tilde{x}tilde{y}+4tilde{y}^2)-10(2tilde{x}+tilde{y})\
&=&5tilde{y}^2-20tilde{x}-10tilde{y}.
end{eqnarray*}
You seem to have made a mistake in your calculation.
answered Jan 21 at 20:50
ServaesServaes
27.5k34098
$endgroup$
add a comment |
$begingroup$
Your approach is entirely correct; plugging in your substitutions into the first equation for $C$ yields the result you claim it should be:
begin{eqnarray*}
x^2+4xy+4y^2-10sqrt{5}x&=&
left(frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}right)^2+4left(frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}right)left(-frac{1}{sqrt{5}}tilde{x}+frac{2}{sqrt{5}}tilde{y}right)\
& &+4left(-frac{1}{sqrt{5}}tilde{x}+frac{2}{sqrt{5}}tilde{y}right)^2-10sqrt{5}left(frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}right)\
&=&frac{1}{5}(4tilde{x}^2+4tilde{x}tilde{y}+tilde{y}^2)+frac{4}{5}(-2tilde{x}^2+3tilde{x}tilde{y}+2tilde{y}^2)\
& &+frac{4}{5}(tilde{x}^2-4tilde{x}tilde{y}+4tilde{y}^2)-10(2tilde{x}+tilde{y})\
&=&5tilde{y}^2-20tilde{x}-10tilde{y}.
end{eqnarray*}
You seem to have made a mistake in your calculation.
answered Jan 21 at 20:50
ServaesServaes
27.5k34098
$endgroup$
Your approach is entirely correct; plugging in your substitutions into the first equation for $C$ yields the result you claim it should be:
begin{eqnarray*}
x^2+4xy+4y^2-10sqrt{5}x&=&
left(frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}right)^2+4left(frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}right)left(-frac{1}{sqrt{5}}tilde{x}+frac{2}{sqrt{5}}tilde{y}right)\
& &+4left(-frac{1}{sqrt{5}}tilde{x}+frac{2}{sqrt{5}}tilde{y}right)^2-10sqrt{5}left(frac{2}{sqrt{5}}tilde{x}+frac{1}{sqrt{5}}tilde{y}right)\
&=&frac{1}{5}(4tilde{x}^2+4tilde{x}tilde{y}+tilde{y}^2)+frac{4}{5}(-2tilde{x}^2+3tilde{x}tilde{y}+2tilde{y}^2)\
& &+frac{4}{5}(tilde{x}^2-4tilde{x}tilde{y}+4tilde{y}^2)-10(2tilde{x}+tilde{y})\
&=&5tilde{y}^2-20tilde{x}-10tilde{y}.
end{eqnarray*}
You seem to have made a mistake in your calculation.
answered Jan 21 at 20:50
ServaesServaes
27.5k34098
edited Jan 21 at 21:01
answered Jan 21 at 20:50
ServaesServaes
27.5k34098
answered Jan 21 at 20:50
ServaesServaes
27.5k34098
answered Jan 21 at 20:50
ServaesServaes
27.5k34098
27.5k34098
add a comment |
add a comment |
$begingroup$
Basically, it looks like you’re getting the transformation backwards. If you have a coordinate transformation $mathbf x'=Qmathbf x$, to get the transformed equation you have to substitute for $mathbf x$ in the original equation, i.e., apply the inverse transformation to $x$ and $y$ in the equation. So, if the original coordinate transformation was given by $Q$, you now want to transform by $Q^{-1}$, which means that you substitute $Q(x,y)^T$ for $(x,y)$ in the equation that you have.
answered Jan 21 at 21:00
amdamd
30.8k21051
$endgroup$
add a comment |
$begingroup$
Basically, it looks like you’re getting the transformation backwards. If you have a coordinate transformation $mathbf x'=Qmathbf x$, to get the transformed equation you have to substitute for $mathbf x$ in the original equation, i.e., apply the inverse transformation to $x$ and $y$ in the equation. So, if the original coordinate transformation was given by $Q$, you now want to transform by $Q^{-1}$, which means that you substitute $Q(x,y)^T$ for $(x,y)$ in the equation that you have.
answered Jan 21 at 21:00
amdamd
30.8k21051
$endgroup$
add a comment |
$begingroup$
Basically, it looks like you’re getting the transformation backwards. If you have a coordinate transformation $mathbf x'=Qmathbf x$, to get the transformed equation you have to substitute for $mathbf x$ in the original equation, i.e., apply the inverse transformation to $x$ and $y$ in the equation. So, if the original coordinate transformation was given by $Q$, you now want to transform by $Q^{-1}$, which means that you substitute $Q(x,y)^T$ for $(x,y)$ in the equation that you have.
answered Jan 21 at 21:00
amdamd
30.8k21051
$endgroup$
Basically, it looks like you’re getting the transformation backwards. If you have a coordinate transformation $mathbf x'=Qmathbf x$, to get the transformed equation you have to substitute for $mathbf x$ in the original equation, i.e., apply the inverse transformation to $x$ and $y$ in the equation. So, if the original coordinate transformation was given by $Q$, you now want to transform by $Q^{-1}$, which means that you substitute $Q(x,y)^T$ for $(x,y)$ in the equation that you have.
answered Jan 21 at 21:00
amdamd
30.8k21051
answered Jan 21 at 21:00
amdamd
30.8k21051
answered Jan 21 at 21:00
amdamd
30.8k21051
answered Jan 21 at 21:00
amdamd
30.8k21051
30.8k21051
add a comment |
add a comment |
$begingroup$
Suppose we have a conic with the equation $P(x,y) = 0.$
Next, you apply the rotation $Q$ to the conic,
so if $(tilde x,tilde y)$ is a point on the conic before rotation,
it is taken to the point
$$
(x,y)=left(tfrac{2}{sqrt5}tilde{x}+tfrac{1}{sqrt5}tilde{y},-tfrac{1}{sqrt5}tilde{x}+tfrac{2}{sqrt5}tilde{y}right) tag1
$$
after rotation.
Moreover, after applying the rotation, the rotated conic has the equation
$$ x^2+4xy+4y^2-10sqrt{5}x=0. tag2$$
If you now use Equation $(1)$ to provide the values of $x$ and $y$
in terms of $tilde x$ and $tilde y,$
you can make those substitutions for $x$ and $y$ in Equation $(2),$
and then you have an equation that must be satisfied by any point
$(tilde x,tilde y)$ on the original conic.
Try it.
The fact that one of your steps was to "solve" Equation $(1)$ suggests that your substitution was actually for a rotation by the correct angular magnitude, but in the opposite direction from what you needed.
answered Jan 21 at 21:16
David KDavid K
54.9k344120
$endgroup$
add a comment |
$begingroup$
Suppose we have a conic with the equation $P(x,y) = 0.$
Next, you apply the rotation $Q$ to the conic,
so if $(tilde x,tilde y)$ is a point on the conic before rotation,
it is taken to the point
$$
(x,y)=left(tfrac{2}{sqrt5}tilde{x}+tfrac{1}{sqrt5}tilde{y},-tfrac{1}{sqrt5}tilde{x}+tfrac{2}{sqrt5}tilde{y}right) tag1
$$
after rotation.
Moreover, after applying the rotation, the rotated conic has the equation
$$ x^2+4xy+4y^2-10sqrt{5}x=0. tag2$$
If you now use Equation $(1)$ to provide the values of $x$ and $y$
in terms of $tilde x$ and $tilde y,$
you can make those substitutions for $x$ and $y$ in Equation $(2),$
and then you have an equation that must be satisfied by any point
$(tilde x,tilde y)$ on the original conic.
Try it.
The fact that one of your steps was to "solve" Equation $(1)$ suggests that your substitution was actually for a rotation by the correct angular magnitude, but in the opposite direction from what you needed.
answered Jan 21 at 21:16
David KDavid K
54.9k344120
$endgroup$
add a comment |
$begingroup$
Suppose we have a conic with the equation $P(x,y) = 0.$
Next, you apply the rotation $Q$ to the conic,
so if $(tilde x,tilde y)$ is a point on the conic before rotation,
it is taken to the point
$$
(x,y)=left(tfrac{2}{sqrt5}tilde{x}+tfrac{1}{sqrt5}tilde{y},-tfrac{1}{sqrt5}tilde{x}+tfrac{2}{sqrt5}tilde{y}right) tag1
$$
after rotation.
Moreover, after applying the rotation, the rotated conic has the equation
$$ x^2+4xy+4y^2-10sqrt{5}x=0. tag2$$
If you now use Equation $(1)$ to provide the values of $x$ and $y$
in terms of $tilde x$ and $tilde y,$
you can make those substitutions for $x$ and $y$ in Equation $(2),$
and then you have an equation that must be satisfied by any point
$(tilde x,tilde y)$ on the original conic.
Try it.
The fact that one of your steps was to "solve" Equation $(1)$ suggests that your substitution was actually for a rotation by the correct angular magnitude, but in the opposite direction from what you needed.
answered Jan 21 at 21:16
David KDavid K
54.9k344120
$endgroup$
Suppose we have a conic with the equation $P(x,y) = 0.$
Next, you apply the rotation $Q$ to the conic,
so if $(tilde x,tilde y)$ is a point on the conic before rotation,
it is taken to the point
$$
(x,y)=left(tfrac{2}{sqrt5}tilde{x}+tfrac{1}{sqrt5}tilde{y},-tfrac{1}{sqrt5}tilde{x}+tfrac{2}{sqrt5}tilde{y}right) tag1
$$
after rotation.
Moreover, after applying the rotation, the rotated conic has the equation
$$ x^2+4xy+4y^2-10sqrt{5}x=0. tag2$$
If you now use Equation $(1)$ to provide the values of $x$ and $y$
in terms of $tilde x$ and $tilde y,$
you can make those substitutions for $x$ and $y$ in Equation $(2),$
and then you have an equation that must be satisfied by any point
$(tilde x,tilde y)$ on the original conic.
Try it.
The fact that one of your steps was to "solve" Equation $(1)$ suggests that your substitution was actually for a rotation by the correct angular magnitude, but in the opposite direction from what you needed.
answered Jan 21 at 21:16
David KDavid K
54.9k344120
answered Jan 21 at 21:16
David KDavid K
54.9k344120
answered Jan 21 at 21:16
David KDavid K
54.9k344120
answered Jan 21 at 21:16
David KDavid K
54.9k344120
54.9k344120
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$begingroup$
You mean you want to find the equation after applying the rotation? Before applying it, you have the equation that you wrote down, or am I misunderstanding?
$endgroup$
– Servaes
Jan 21 at 20:48
1
$begingroup$
I'm not sure but you may be applying the inverse rotation.
$endgroup$
– Harnak
Jan 21 at 20:49
$begingroup$
Another way, substantially equivalent but slighty easier in calculations, is to consider $(x,y)^tA(x,y)+Bcdot(x,y)$, and then applying $(tilde{x},tilde{y})=Q(x,y)$
$endgroup$
– gabriele cassese
Jan 21 at 20:49
$begingroup$
@Servaes No, the equation of the conic has been rotated by the rotation matrix. Now, I want to find the equation of the conic before the rotation.
$endgroup$
– Kevin
Jan 21 at 20:50
$begingroup$
@amd $-10sqrt{5}x$
$endgroup$
– gabriele cassese
Jan 21 at 20:51