Mixing
Equation of auxiliary circle of the ellipse $2x^2 +6xy + 5y^2$ =1
$begingroup$
Equation of auxiliary circle of the ellipse $2x^2 +6xy + 5y^2$= 1
My approach is , First I try remove xy term from the equation, to convert the given equation in the standard equation of ellipse and find the value of $a$ and $b$. For this , I use the concept of rotation of axis, using this concept first I find $tan2theta = -2$ then I find the value of $sin theta$ and $cos theta$. But the problem is the the value of $sin theta$ and $costheta$ are complex and when I try to solve using the concept of rotation of axis but equation become too much complex and also take a lots of time to solve it. So I please tell me another approach or method to solve this question.
conic-sections
asked Jan 11 at 19:22
saket kumarsaket kumar
586
$endgroup$
add a comment |
$begingroup$
Equation of auxiliary circle of the ellipse $2x^2 +6xy + 5y^2$= 1
My approach is , First I try remove xy term from the equation, to convert the given equation in the standard equation of ellipse and find the value of $a$ and $b$. For this , I use the concept of rotation of axis, using this concept first I find $tan2theta = -2$ then I find the value of $sin theta$ and $cos theta$. But the problem is the the value of $sin theta$ and $costheta$ are complex and when I try to solve using the concept of rotation of axis but equation become too much complex and also take a lots of time to solve it. So I please tell me another approach or method to solve this question.
conic-sections
asked Jan 11 at 19:22
saket kumarsaket kumar
586
$endgroup$
$begingroup$
First of all $2x^2+6xy+5y^2$ is not an equation. That aside, apply a rotation with unknown sine and cosine to this and use the fact that the sum of their squares is equal to one to find the correct values to eliminate the cross term. If the resulting expressions are complicated, you’ll just have to deal with them. A simpler approach is to compute the eigenvalues of this quadratic form, but those expressions will also involve radicals.
$endgroup$
– amd
Jan 11 at 20:56
$begingroup$
Ok I try to solve the resulting expression.
$endgroup$
– saket kumar
Jan 12 at 3:28
add a comment |
$begingroup$
Equation of auxiliary circle of the ellipse $2x^2 +6xy + 5y^2$= 1
My approach is , First I try remove xy term from the equation, to convert the given equation in the standard equation of ellipse and find the value of $a$ and $b$. For this , I use the concept of rotation of axis, using this concept first I find $tan2theta = -2$ then I find the value of $sin theta$ and $cos theta$. But the problem is the the value of $sin theta$ and $costheta$ are complex and when I try to solve using the concept of rotation of axis but equation become too much complex and also take a lots of time to solve it. So I please tell me another approach or method to solve this question.
conic-sections
asked Jan 11 at 19:22
saket kumarsaket kumar
586
$endgroup$
Equation of auxiliary circle of the ellipse $2x^2 +6xy + 5y^2$= 1
My approach is , First I try remove xy term from the equation, to convert the given equation in the standard equation of ellipse and find the value of $a$ and $b$. For this , I use the concept of rotation of axis, using this concept first I find $tan2theta = -2$ then I find the value of $sin theta$ and $cos theta$. But the problem is the the value of $sin theta$ and $costheta$ are complex and when I try to solve using the concept of rotation of axis but equation become too much complex and also take a lots of time to solve it. So I please tell me another approach or method to solve this question.
conic-sections
conic-sections
asked Jan 11 at 19:22
saket kumarsaket kumar
586
asked Jan 11 at 19:22
saket kumarsaket kumar
586
edited Jan 12 at 2:40
saket kumar
asked Jan 11 at 19:22
saket kumarsaket kumar
586
asked Jan 11 at 19:22
saket kumarsaket kumar
586
asked Jan 11 at 19:22
saket kumarsaket kumar
586
586
$begingroup$
First of all $2x^2+6xy+5y^2$ is not an equation. That aside, apply a rotation with unknown sine and cosine to this and use the fact that the sum of their squares is equal to one to find the correct values to eliminate the cross term. If the resulting expressions are complicated, you’ll just have to deal with them. A simpler approach is to compute the eigenvalues of this quadratic form, but those expressions will also involve radicals.
$endgroup$
– amd
Jan 11 at 20:56
$begingroup$
Ok I try to solve the resulting expression.
$endgroup$
– saket kumar
Jan 12 at 3:28
add a comment |
$begingroup$
First of all $2x^2+6xy+5y^2$ is not an equation. That aside, apply a rotation with unknown sine and cosine to this and use the fact that the sum of their squares is equal to one to find the correct values to eliminate the cross term. If the resulting expressions are complicated, you’ll just have to deal with them. A simpler approach is to compute the eigenvalues of this quadratic form, but those expressions will also involve radicals.
$endgroup$
– amd
Jan 11 at 20:56
$begingroup$
Ok I try to solve the resulting expression.
$endgroup$
– saket kumar
Jan 12 at 3:28
$begingroup$
First of all $2x^2+6xy+5y^2$ is not an equation. That aside, apply a rotation with unknown sine and cosine to this and use the fact that the sum of their squares is equal to one to find the correct values to eliminate the cross term. If the resulting expressions are complicated, you’ll just have to deal with them. A simpler approach is to compute the eigenvalues of this quadratic form, but those expressions will also involve radicals.
$endgroup$
– amd
Jan 11 at 20:56
$begingroup$
First of all $2x^2+6xy+5y^2$ is not an equation. That aside, apply a rotation with unknown sine and cosine to this and use the fact that the sum of their squares is equal to one to find the correct values to eliminate the cross term. If the resulting expressions are complicated, you’ll just have to deal with them. A simpler approach is to compute the eigenvalues of this quadratic form, but those expressions will also involve radicals.
$endgroup$
– amd
Jan 11 at 20:56
$begingroup$
Ok I try to solve the resulting expression.
$endgroup$
– saket kumar
Jan 12 at 3:28
$begingroup$
Ok I try to solve the resulting expression.
$endgroup$
– saket kumar
Jan 12 at 3:28
add a comment |
2 Answers
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The auxiliary circle is centered at the center of the ellipse and its radius equals the semi-major axis of the ellipse. Any conic of the form $ax^2+2hxy+by^2=1$ is centered at origin (as if $(x,y)$ lies on the ellipse so does $(-x,-y)$).
To find the semi-major axis we need to find the greatest distance of any point on the conic from the center. So we turn to polar coordinates $x=r cos theta, y= rsin theta$
Substituting in the equation we get $r^2 = dfrac{1}{2 cos^2 theta+6 sin theta cos theta+ 5 sin^2 theta} = dfrac{2}{7+6sin 2 theta - 3 cos 2 theta} le dfrac{7+3 sqrt 5}{2}$
Hence the equation of the auxiliary circle is $x^2+y^2 = dfrac{7+3 sqrt 5}{2}$
answered Jan 12 at 11:23
Hari ShankarHari Shankar
2,276139
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$begingroup$
This is good. I was thinking of a more tedious approach.
$endgroup$
– Lubin
Jan 13 at 4:50
add a comment |
$begingroup$
To avoid rotations, you could find the intersection $P$ between the ellipse (which is centred at $O=(0,0)$) and a generic line through the centre $y=mx$. Length $PO$ is then an expression in $m$ whose maximum gives the radius of the auxiliary circle.
answered Jan 12 at 11:27
AretinoAretino
23.6k21443
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
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active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
The auxiliary circle is centered at the center of the ellipse and its radius equals the semi-major axis of the ellipse. Any conic of the form $ax^2+2hxy+by^2=1$ is centered at origin (as if $(x,y)$ lies on the ellipse so does $(-x,-y)$).
To find the semi-major axis we need to find the greatest distance of any point on the conic from the center. So we turn to polar coordinates $x=r cos theta, y= rsin theta$
Substituting in the equation we get $r^2 = dfrac{1}{2 cos^2 theta+6 sin theta cos theta+ 5 sin^2 theta} = dfrac{2}{7+6sin 2 theta - 3 cos 2 theta} le dfrac{7+3 sqrt 5}{2}$
Hence the equation of the auxiliary circle is $x^2+y^2 = dfrac{7+3 sqrt 5}{2}$
answered Jan 12 at 11:23
Hari ShankarHari Shankar
2,276139
$endgroup$
$begingroup$
This is good. I was thinking of a more tedious approach.
$endgroup$
– Lubin
Jan 13 at 4:50
add a comment |
$begingroup$
The auxiliary circle is centered at the center of the ellipse and its radius equals the semi-major axis of the ellipse. Any conic of the form $ax^2+2hxy+by^2=1$ is centered at origin (as if $(x,y)$ lies on the ellipse so does $(-x,-y)$).
To find the semi-major axis we need to find the greatest distance of any point on the conic from the center. So we turn to polar coordinates $x=r cos theta, y= rsin theta$
Substituting in the equation we get $r^2 = dfrac{1}{2 cos^2 theta+6 sin theta cos theta+ 5 sin^2 theta} = dfrac{2}{7+6sin 2 theta - 3 cos 2 theta} le dfrac{7+3 sqrt 5}{2}$
Hence the equation of the auxiliary circle is $x^2+y^2 = dfrac{7+3 sqrt 5}{2}$
answered Jan 12 at 11:23
Hari ShankarHari Shankar
2,276139
$endgroup$
$begingroup$
This is good. I was thinking of a more tedious approach.
$endgroup$
– Lubin
Jan 13 at 4:50
add a comment |
$begingroup$
The auxiliary circle is centered at the center of the ellipse and its radius equals the semi-major axis of the ellipse. Any conic of the form $ax^2+2hxy+by^2=1$ is centered at origin (as if $(x,y)$ lies on the ellipse so does $(-x,-y)$).
To find the semi-major axis we need to find the greatest distance of any point on the conic from the center. So we turn to polar coordinates $x=r cos theta, y= rsin theta$
Substituting in the equation we get $r^2 = dfrac{1}{2 cos^2 theta+6 sin theta cos theta+ 5 sin^2 theta} = dfrac{2}{7+6sin 2 theta - 3 cos 2 theta} le dfrac{7+3 sqrt 5}{2}$
Hence the equation of the auxiliary circle is $x^2+y^2 = dfrac{7+3 sqrt 5}{2}$
answered Jan 12 at 11:23
Hari ShankarHari Shankar
2,276139
$endgroup$
The auxiliary circle is centered at the center of the ellipse and its radius equals the semi-major axis of the ellipse. Any conic of the form $ax^2+2hxy+by^2=1$ is centered at origin (as if $(x,y)$ lies on the ellipse so does $(-x,-y)$).
To find the semi-major axis we need to find the greatest distance of any point on the conic from the center. So we turn to polar coordinates $x=r cos theta, y= rsin theta$
Substituting in the equation we get $r^2 = dfrac{1}{2 cos^2 theta+6 sin theta cos theta+ 5 sin^2 theta} = dfrac{2}{7+6sin 2 theta - 3 cos 2 theta} le dfrac{7+3 sqrt 5}{2}$
Hence the equation of the auxiliary circle is $x^2+y^2 = dfrac{7+3 sqrt 5}{2}$
answered Jan 12 at 11:23
Hari ShankarHari Shankar
2,276139
answered Jan 12 at 11:23
Hari ShankarHari Shankar
2,276139
answered Jan 12 at 11:23
Hari ShankarHari Shankar
2,276139
answered Jan 12 at 11:23
Hari ShankarHari Shankar
2,276139
2,276139
$begingroup$
This is good. I was thinking of a more tedious approach.
$endgroup$
– Lubin
Jan 13 at 4:50
add a comment |
$begingroup$
This is good. I was thinking of a more tedious approach.
$endgroup$
– Lubin
Jan 13 at 4:50
$begingroup$
This is good. I was thinking of a more tedious approach.
$endgroup$
– Lubin
Jan 13 at 4:50
$begingroup$
This is good. I was thinking of a more tedious approach.
$endgroup$
– Lubin
Jan 13 at 4:50
add a comment |
$begingroup$
To avoid rotations, you could find the intersection $P$ between the ellipse (which is centred at $O=(0,0)$) and a generic line through the centre $y=mx$. Length $PO$ is then an expression in $m$ whose maximum gives the radius of the auxiliary circle.
answered Jan 12 at 11:27
AretinoAretino
23.6k21443
$endgroup$
add a comment |
$begingroup$
To avoid rotations, you could find the intersection $P$ between the ellipse (which is centred at $O=(0,0)$) and a generic line through the centre $y=mx$. Length $PO$ is then an expression in $m$ whose maximum gives the radius of the auxiliary circle.
answered Jan 12 at 11:27
AretinoAretino
23.6k21443
$endgroup$
add a comment |
$begingroup$
To avoid rotations, you could find the intersection $P$ between the ellipse (which is centred at $O=(0,0)$) and a generic line through the centre $y=mx$. Length $PO$ is then an expression in $m$ whose maximum gives the radius of the auxiliary circle.
answered Jan 12 at 11:27
AretinoAretino
23.6k21443
$endgroup$
To avoid rotations, you could find the intersection $P$ between the ellipse (which is centred at $O=(0,0)$) and a generic line through the centre $y=mx$. Length $PO$ is then an expression in $m$ whose maximum gives the radius of the auxiliary circle.
answered Jan 12 at 11:27
AretinoAretino
23.6k21443
answered Jan 12 at 11:27
AretinoAretino
23.6k21443
answered Jan 12 at 11:27
AretinoAretino
23.6k21443
answered Jan 12 at 11:27
AretinoAretino
23.6k21443
23.6k21443
add a comment |
add a comment |
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$begingroup$
First of all $2x^2+6xy+5y^2$ is not an equation. That aside, apply a rotation with unknown sine and cosine to this and use the fact that the sum of their squares is equal to one to find the correct values to eliminate the cross term. If the resulting expressions are complicated, you’ll just have to deal with them. A simpler approach is to compute the eigenvalues of this quadratic form, but those expressions will also involve radicals.
$endgroup$
– amd
Jan 11 at 20:56
$begingroup$
Ok I try to solve the resulting expression.
$endgroup$
– saket kumar
Jan 12 at 3:28