Mixing
How to determine the area of a rotated ellipse?
$begingroup$
The ellipse $6x^2+4xy+5y^2+8x+8y+1=0$ is neither expressed in terms of $x$; like $y=pmsqrt{a^2-x^2}$, nor in terms of $y$; like $x=pmsqrt{a^2-y^2}$.
Separation of $x$ (or $y%$) may be impossible.
I was thinking to let $x=rcos(theta)$, and $y=rsin(theta)$, and then integrating, but finding the limits of integration [with respect to $theta$] is difficult, or may be impossible.
How to determine the area of the ellipse shown below?
integration conic-sections area polar-coordinates
asked Jan 13 at 12:15
Hussain-AlqatariHussain-Alqatari
3267
$endgroup$
add a comment |
$begingroup$
The ellipse $6x^2+4xy+5y^2+8x+8y+1=0$ is neither expressed in terms of $x$; like $y=pmsqrt{a^2-x^2}$, nor in terms of $y$; like $x=pmsqrt{a^2-y^2}$.
Separation of $x$ (or $y%$) may be impossible.
I was thinking to let $x=rcos(theta)$, and $y=rsin(theta)$, and then integrating, but finding the limits of integration [with respect to $theta$] is difficult, or may be impossible.
How to determine the area of the ellipse shown below?
integration conic-sections area polar-coordinates
asked Jan 13 at 12:15
Hussain-AlqatariHussain-Alqatari
3267
$endgroup$
1
$begingroup$
Hint: Complete the Squares, Shift the centre to the origin and then convert to polar coordinates and integrate
$endgroup$
– DavidG
Jan 13 at 12:32
1
$begingroup$
Following David's comment, if you know how to classify quadratics with matrices and etc., say, you can use that to "make" your ellipse a canonical one (i.e. with center at the origin) and then integrate there.
$endgroup$
– DonAntonio
Jan 13 at 12:34
$begingroup$
@DavidG, when completing the squares, the term $4xy$ disturbs. I have reached to $6(x+frac{2}{3})^2+5(y+frac{4}{5})^2=-4xy+frac{73}{15}$
$endgroup$
– Hussain-Alqatari
Jan 13 at 13:12
add a comment |
$begingroup$
The ellipse $6x^2+4xy+5y^2+8x+8y+1=0$ is neither expressed in terms of $x$; like $y=pmsqrt{a^2-x^2}$, nor in terms of $y$; like $x=pmsqrt{a^2-y^2}$.
Separation of $x$ (or $y%$) may be impossible.
I was thinking to let $x=rcos(theta)$, and $y=rsin(theta)$, and then integrating, but finding the limits of integration [with respect to $theta$] is difficult, or may be impossible.
How to determine the area of the ellipse shown below?
integration conic-sections area polar-coordinates
asked Jan 13 at 12:15
Hussain-AlqatariHussain-Alqatari
3267
$endgroup$
The ellipse $6x^2+4xy+5y^2+8x+8y+1=0$ is neither expressed in terms of $x$; like $y=pmsqrt{a^2-x^2}$, nor in terms of $y$; like $x=pmsqrt{a^2-y^2}$.
Separation of $x$ (or $y%$) may be impossible.
I was thinking to let $x=rcos(theta)$, and $y=rsin(theta)$, and then integrating, but finding the limits of integration [with respect to $theta$] is difficult, or may be impossible.
How to determine the area of the ellipse shown below?
integration conic-sections area polar-coordinates
integration conic-sections area polar-coordinates
asked Jan 13 at 12:15
Hussain-AlqatariHussain-Alqatari
3267
asked Jan 13 at 12:15
Hussain-AlqatariHussain-Alqatari
3267
asked Jan 13 at 12:15
Hussain-AlqatariHussain-Alqatari
3267
asked Jan 13 at 12:15
Hussain-AlqatariHussain-Alqatari
3267
asked Jan 13 at 12:15
Hussain-AlqatariHussain-Alqatari
3267
3267
1
$begingroup$
Hint: Complete the Squares, Shift the centre to the origin and then convert to polar coordinates and integrate
$endgroup$
– DavidG
Jan 13 at 12:32
1
$begingroup$
Following David's comment, if you know how to classify quadratics with matrices and etc., say, you can use that to "make" your ellipse a canonical one (i.e. with center at the origin) and then integrate there.
$endgroup$
– DonAntonio
Jan 13 at 12:34
$begingroup$
@DavidG, when completing the squares, the term $4xy$ disturbs. I have reached to $6(x+frac{2}{3})^2+5(y+frac{4}{5})^2=-4xy+frac{73}{15}$
$endgroup$
– Hussain-Alqatari
Jan 13 at 13:12
add a comment |
1
$begingroup$
Hint: Complete the Squares, Shift the centre to the origin and then convert to polar coordinates and integrate
$endgroup$
– DavidG
Jan 13 at 12:32
1
$begingroup$
Following David's comment, if you know how to classify quadratics with matrices and etc., say, you can use that to "make" your ellipse a canonical one (i.e. with center at the origin) and then integrate there.
$endgroup$
– DonAntonio
Jan 13 at 12:34
$begingroup$
@DavidG, when completing the squares, the term $4xy$ disturbs. I have reached to $6(x+frac{2}{3})^2+5(y+frac{4}{5})^2=-4xy+frac{73}{15}$
$endgroup$
– Hussain-Alqatari
Jan 13 at 13:12
1
1
$begingroup$
Hint: Complete the Squares, Shift the centre to the origin and then convert to polar coordinates and integrate
$endgroup$
– DavidG
Jan 13 at 12:32
$begingroup$
Hint: Complete the Squares, Shift the centre to the origin and then convert to polar coordinates and integrate
$endgroup$
– DavidG
Jan 13 at 12:32
1
1
$begingroup$
Following David's comment, if you know how to classify quadratics with matrices and etc., say, you can use that to "make" your ellipse a canonical one (i.e. with center at the origin) and then integrate there.
$endgroup$
– DonAntonio
Jan 13 at 12:34
$begingroup$
Following David's comment, if you know how to classify quadratics with matrices and etc., say, you can use that to "make" your ellipse a canonical one (i.e. with center at the origin) and then integrate there.
$endgroup$
– DonAntonio
Jan 13 at 12:34
$begingroup$
@DavidG, when completing the squares, the term $4xy$ disturbs. I have reached to $6(x+frac{2}{3})^2+5(y+frac{4}{5})^2=-4xy+frac{73}{15}$
$endgroup$
– Hussain-Alqatari
Jan 13 at 13:12
$begingroup$
@DavidG, when completing the squares, the term $4xy$ disturbs. I have reached to $6(x+frac{2}{3})^2+5(y+frac{4}{5})^2=-4xy+frac{73}{15}$
$endgroup$
– Hussain-Alqatari
Jan 13 at 13:12
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Consider the ellipse $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$. It is well known that (1) this conic is an ellipse iff $AB - H^2 > 0$, so that we can assume $A > 0$; and (2) assuming $A > 0,$ the ellipse has real points iff $L > 0$, where
$$L = AF^2 + BG^2 + CH^2 - ABC - 2FGH.$$
To find area of the ellipse, one can apply an affine transformation that doesn't change the area. Such a transformation is $x = lambda y + z$ where $lambda$ is any constant.
This gives
$$(Alambda^2 + 2Hlambda + B)y^2 + 2(Alambda + H)yz + Az^2 +2(Glambda + F)y + 2Gz + C = 0.$$
Get rid of $yz$ by choosing $lambda = -H/A.$ Then
$$(AB - H^2)y^2 + 2(FA - GH)y + A^2z^2 + 2GAz + AC = 0,$$
that is
$$(AB - H^2)left(y - {{(AF - GH)}over{AB - H^2}}right)^2 + (Az + G)^2 - {{(AF - GH)^2}over{AB - H^2}} - G^2 + AC = 0.$$
Moving the centre of the ellipse to the origin, this can be put as
$$(AB - H^2){y_1}^2 + A^2{z_1}^2 - K = 0,$$
where
$$K = {1over{(AB - H^2)}}left[ (AF - GH)^2 + (AB - H^2)(G^2 - AC)right],$$
which reduces to $K = LA/(AB - H^2)$ (with $L$ as above). Hence the semi-axes of the transformed ellipse are
$a = sqrt{K/(AB - H^2)}$ and $b = sqrt{K}/A,$ and the required area is $pi ab = pi L/(AB - H^2)^{3/2}.$
In particular, the ellipse in the question has area $86pi/26^{3/2}.$
answered Jan 13 at 16:49
Michael BehrendMichael Behrend
1,22746
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add a comment |
$begingroup$
Write the equation as:
$$begin{pmatrix}x & y & 1end{pmatrix}
left(begin{array}{ccc|c} \&Q& &P \ \ hline &P^T& &R end{array}right)
begin{pmatrix}x \ y \ 1end{pmatrix}
=begin{pmatrix}x & y & 1end{pmatrix}
left(begin{array}{cc|c}6 & 2 & 4 \ 2 & 5 & 4 \ hline 4&4&1 end{array}right)
begin{pmatrix}x \ y \ 1end{pmatrix} = 0$$
Set the gradient equal to zero to find the translation (same as solving $Qx=-P$):
$$nabla(6x^2+4xy+5y^2+8x+8y+1)=0 Rightarrow begin{cases}12x+4y+8=0\4x+10y+8=0end{cases} Rightarrow x=-frac 6{13}, y=-frac 8{13}$$
So the translation vector $t=-frac 1{13}begin{pmatrix}6\8end{pmatrix}$.
Divide the equation by $-(Pcdot t + R)$ to normalize it. The resulting $Q'$ is:
$$Q'=-frac{1}{Pcdot t + R}Q$$
Its determinant is the product of the eigenvalues, which are $frac 1{a^2}$ and $frac 1{b^2}$.
So:
$$det Q' = frac{1}{a^2b^2}$$
The area of an ellipse is $pi a b$ so that:
$$text{Area} = pi a b = frac{pi}{sqrt{det Q'}} = frac{picdot |Pcdot t+R|}{sqrt{|det Q|}}
=frac{picdot |-frac 1{13}(4cdot 6 + 4cdot 8)+1|}{sqrt{|6cdot 5-2cdot 2|}}
=frac{43pi}{13sqrt{26}}$$
The general formula is:
$$text{Area}=frac{pi|P^TQ^{-1}P-R|}{sqrt{|det Q|}}$$
answered Jan 14 at 20:43
I like SerenaI like Serena
4,2071722
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add a comment |
$begingroup$
The simplest way to compute the area is that of using the formulas given here to find semi-axes $a$ and $b$ and then area $=pi ab$.
If you insist on using an integral, you can of course solve for $y$ to find two solutions
$$
y_text{up/down}={1over5}left(
-2x-4pmsqrt{-26x^2-24x+3}
right)
$$
and then compute the area as
$$
int_{x_text{left}}^{x_text{right}}(y_{up}-y_{down}),dx,
$$
where $x_text{left/right}$ are the values of $x$ giving a vanishing square root.
answered Jan 13 at 14:18
AretinoAretino
23.6k21443
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the ellipse $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$. It is well known that (1) this conic is an ellipse iff $AB - H^2 > 0$, so that we can assume $A > 0$; and (2) assuming $A > 0,$ the ellipse has real points iff $L > 0$, where
$$L = AF^2 + BG^2 + CH^2 - ABC - 2FGH.$$
To find area of the ellipse, one can apply an affine transformation that doesn't change the area. Such a transformation is $x = lambda y + z$ where $lambda$ is any constant.
This gives
$$(Alambda^2 + 2Hlambda + B)y^2 + 2(Alambda + H)yz + Az^2 +2(Glambda + F)y + 2Gz + C = 0.$$
Get rid of $yz$ by choosing $lambda = -H/A.$ Then
$$(AB - H^2)y^2 + 2(FA - GH)y + A^2z^2 + 2GAz + AC = 0,$$
that is
$$(AB - H^2)left(y - {{(AF - GH)}over{AB - H^2}}right)^2 + (Az + G)^2 - {{(AF - GH)^2}over{AB - H^2}} - G^2 + AC = 0.$$
Moving the centre of the ellipse to the origin, this can be put as
$$(AB - H^2){y_1}^2 + A^2{z_1}^2 - K = 0,$$
where
$$K = {1over{(AB - H^2)}}left[ (AF - GH)^2 + (AB - H^2)(G^2 - AC)right],$$
which reduces to $K = LA/(AB - H^2)$ (with $L$ as above). Hence the semi-axes of the transformed ellipse are
$a = sqrt{K/(AB - H^2)}$ and $b = sqrt{K}/A,$ and the required area is $pi ab = pi L/(AB - H^2)^{3/2}.$
In particular, the ellipse in the question has area $86pi/26^{3/2}.$
answered Jan 13 at 16:49
Michael BehrendMichael Behrend
1,22746
$endgroup$
add a comment |
$begingroup$
Consider the ellipse $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$. It is well known that (1) this conic is an ellipse iff $AB - H^2 > 0$, so that we can assume $A > 0$; and (2) assuming $A > 0,$ the ellipse has real points iff $L > 0$, where
$$L = AF^2 + BG^2 + CH^2 - ABC - 2FGH.$$
To find area of the ellipse, one can apply an affine transformation that doesn't change the area. Such a transformation is $x = lambda y + z$ where $lambda$ is any constant.
This gives
$$(Alambda^2 + 2Hlambda + B)y^2 + 2(Alambda + H)yz + Az^2 +2(Glambda + F)y + 2Gz + C = 0.$$
Get rid of $yz$ by choosing $lambda = -H/A.$ Then
$$(AB - H^2)y^2 + 2(FA - GH)y + A^2z^2 + 2GAz + AC = 0,$$
that is
$$(AB - H^2)left(y - {{(AF - GH)}over{AB - H^2}}right)^2 + (Az + G)^2 - {{(AF - GH)^2}over{AB - H^2}} - G^2 + AC = 0.$$
Moving the centre of the ellipse to the origin, this can be put as
$$(AB - H^2){y_1}^2 + A^2{z_1}^2 - K = 0,$$
where
$$K = {1over{(AB - H^2)}}left[ (AF - GH)^2 + (AB - H^2)(G^2 - AC)right],$$
which reduces to $K = LA/(AB - H^2)$ (with $L$ as above). Hence the semi-axes of the transformed ellipse are
$a = sqrt{K/(AB - H^2)}$ and $b = sqrt{K}/A,$ and the required area is $pi ab = pi L/(AB - H^2)^{3/2}.$
In particular, the ellipse in the question has area $86pi/26^{3/2}.$
answered Jan 13 at 16:49
Michael BehrendMichael Behrend
1,22746
$endgroup$
add a comment |
$begingroup$
Consider the ellipse $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$. It is well known that (1) this conic is an ellipse iff $AB - H^2 > 0$, so that we can assume $A > 0$; and (2) assuming $A > 0,$ the ellipse has real points iff $L > 0$, where
$$L = AF^2 + BG^2 + CH^2 - ABC - 2FGH.$$
To find area of the ellipse, one can apply an affine transformation that doesn't change the area. Such a transformation is $x = lambda y + z$ where $lambda$ is any constant.
This gives
$$(Alambda^2 + 2Hlambda + B)y^2 + 2(Alambda + H)yz + Az^2 +2(Glambda + F)y + 2Gz + C = 0.$$
Get rid of $yz$ by choosing $lambda = -H/A.$ Then
$$(AB - H^2)y^2 + 2(FA - GH)y + A^2z^2 + 2GAz + AC = 0,$$
that is
$$(AB - H^2)left(y - {{(AF - GH)}over{AB - H^2}}right)^2 + (Az + G)^2 - {{(AF - GH)^2}over{AB - H^2}} - G^2 + AC = 0.$$
Moving the centre of the ellipse to the origin, this can be put as
$$(AB - H^2){y_1}^2 + A^2{z_1}^2 - K = 0,$$
where
$$K = {1over{(AB - H^2)}}left[ (AF - GH)^2 + (AB - H^2)(G^2 - AC)right],$$
which reduces to $K = LA/(AB - H^2)$ (with $L$ as above). Hence the semi-axes of the transformed ellipse are
$a = sqrt{K/(AB - H^2)}$ and $b = sqrt{K}/A,$ and the required area is $pi ab = pi L/(AB - H^2)^{3/2}.$
In particular, the ellipse in the question has area $86pi/26^{3/2}.$
answered Jan 13 at 16:49
Michael BehrendMichael Behrend
1,22746
$endgroup$
Consider the ellipse $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$. It is well known that (1) this conic is an ellipse iff $AB - H^2 > 0$, so that we can assume $A > 0$; and (2) assuming $A > 0,$ the ellipse has real points iff $L > 0$, where
$$L = AF^2 + BG^2 + CH^2 - ABC - 2FGH.$$
To find area of the ellipse, one can apply an affine transformation that doesn't change the area. Such a transformation is $x = lambda y + z$ where $lambda$ is any constant.
This gives
$$(Alambda^2 + 2Hlambda + B)y^2 + 2(Alambda + H)yz + Az^2 +2(Glambda + F)y + 2Gz + C = 0.$$
Get rid of $yz$ by choosing $lambda = -H/A.$ Then
$$(AB - H^2)y^2 + 2(FA - GH)y + A^2z^2 + 2GAz + AC = 0,$$
that is
$$(AB - H^2)left(y - {{(AF - GH)}over{AB - H^2}}right)^2 + (Az + G)^2 - {{(AF - GH)^2}over{AB - H^2}} - G^2 + AC = 0.$$
Moving the centre of the ellipse to the origin, this can be put as
$$(AB - H^2){y_1}^2 + A^2{z_1}^2 - K = 0,$$
where
$$K = {1over{(AB - H^2)}}left[ (AF - GH)^2 + (AB - H^2)(G^2 - AC)right],$$
which reduces to $K = LA/(AB - H^2)$ (with $L$ as above). Hence the semi-axes of the transformed ellipse are
$a = sqrt{K/(AB - H^2)}$ and $b = sqrt{K}/A,$ and the required area is $pi ab = pi L/(AB - H^2)^{3/2}.$
In particular, the ellipse in the question has area $86pi/26^{3/2}.$
answered Jan 13 at 16:49
Michael BehrendMichael Behrend
1,22746
answered Jan 13 at 16:49
Michael BehrendMichael Behrend
1,22746
answered Jan 13 at 16:49
Michael BehrendMichael Behrend
1,22746
answered Jan 13 at 16:49
Michael BehrendMichael Behrend
1,22746
1,22746
add a comment |
add a comment |
$begingroup$
Write the equation as:
$$begin{pmatrix}x & y & 1end{pmatrix}
left(begin{array}{ccc|c} \&Q& &P \ \ hline &P^T& &R end{array}right)
begin{pmatrix}x \ y \ 1end{pmatrix}
=begin{pmatrix}x & y & 1end{pmatrix}
left(begin{array}{cc|c}6 & 2 & 4 \ 2 & 5 & 4 \ hline 4&4&1 end{array}right)
begin{pmatrix}x \ y \ 1end{pmatrix} = 0$$
Set the gradient equal to zero to find the translation (same as solving $Qx=-P$):
$$nabla(6x^2+4xy+5y^2+8x+8y+1)=0 Rightarrow begin{cases}12x+4y+8=0\4x+10y+8=0end{cases} Rightarrow x=-frac 6{13}, y=-frac 8{13}$$
So the translation vector $t=-frac 1{13}begin{pmatrix}6\8end{pmatrix}$.
Divide the equation by $-(Pcdot t + R)$ to normalize it. The resulting $Q'$ is:
$$Q'=-frac{1}{Pcdot t + R}Q$$
Its determinant is the product of the eigenvalues, which are $frac 1{a^2}$ and $frac 1{b^2}$.
So:
$$det Q' = frac{1}{a^2b^2}$$
The area of an ellipse is $pi a b$ so that:
$$text{Area} = pi a b = frac{pi}{sqrt{det Q'}} = frac{picdot |Pcdot t+R|}{sqrt{|det Q|}}
=frac{picdot |-frac 1{13}(4cdot 6 + 4cdot 8)+1|}{sqrt{|6cdot 5-2cdot 2|}}
=frac{43pi}{13sqrt{26}}$$
The general formula is:
$$text{Area}=frac{pi|P^TQ^{-1}P-R|}{sqrt{|det Q|}}$$
answered Jan 14 at 20:43
I like SerenaI like Serena
4,2071722
$endgroup$
add a comment |
$begingroup$
Write the equation as:
$$begin{pmatrix}x & y & 1end{pmatrix}
left(begin{array}{ccc|c} \&Q& &P \ \ hline &P^T& &R end{array}right)
begin{pmatrix}x \ y \ 1end{pmatrix}
=begin{pmatrix}x & y & 1end{pmatrix}
left(begin{array}{cc|c}6 & 2 & 4 \ 2 & 5 & 4 \ hline 4&4&1 end{array}right)
begin{pmatrix}x \ y \ 1end{pmatrix} = 0$$
Set the gradient equal to zero to find the translation (same as solving $Qx=-P$):
$$nabla(6x^2+4xy+5y^2+8x+8y+1)=0 Rightarrow begin{cases}12x+4y+8=0\4x+10y+8=0end{cases} Rightarrow x=-frac 6{13}, y=-frac 8{13}$$
So the translation vector $t=-frac 1{13}begin{pmatrix}6\8end{pmatrix}$.
Divide the equation by $-(Pcdot t + R)$ to normalize it. The resulting $Q'$ is:
$$Q'=-frac{1}{Pcdot t + R}Q$$
Its determinant is the product of the eigenvalues, which are $frac 1{a^2}$ and $frac 1{b^2}$.
So:
$$det Q' = frac{1}{a^2b^2}$$
The area of an ellipse is $pi a b$ so that:
$$text{Area} = pi a b = frac{pi}{sqrt{det Q'}} = frac{picdot |Pcdot t+R|}{sqrt{|det Q|}}
=frac{picdot |-frac 1{13}(4cdot 6 + 4cdot 8)+1|}{sqrt{|6cdot 5-2cdot 2|}}
=frac{43pi}{13sqrt{26}}$$
The general formula is:
$$text{Area}=frac{pi|P^TQ^{-1}P-R|}{sqrt{|det Q|}}$$
answered Jan 14 at 20:43
I like SerenaI like Serena
4,2071722
$endgroup$
add a comment |
$begingroup$
Write the equation as:
$$begin{pmatrix}x & y & 1end{pmatrix}
left(begin{array}{ccc|c} \&Q& &P \ \ hline &P^T& &R end{array}right)
begin{pmatrix}x \ y \ 1end{pmatrix}
=begin{pmatrix}x & y & 1end{pmatrix}
left(begin{array}{cc|c}6 & 2 & 4 \ 2 & 5 & 4 \ hline 4&4&1 end{array}right)
begin{pmatrix}x \ y \ 1end{pmatrix} = 0$$
Set the gradient equal to zero to find the translation (same as solving $Qx=-P$):
$$nabla(6x^2+4xy+5y^2+8x+8y+1)=0 Rightarrow begin{cases}12x+4y+8=0\4x+10y+8=0end{cases} Rightarrow x=-frac 6{13}, y=-frac 8{13}$$
So the translation vector $t=-frac 1{13}begin{pmatrix}6\8end{pmatrix}$.
Divide the equation by $-(Pcdot t + R)$ to normalize it. The resulting $Q'$ is:
$$Q'=-frac{1}{Pcdot t + R}Q$$
Its determinant is the product of the eigenvalues, which are $frac 1{a^2}$ and $frac 1{b^2}$.
So:
$$det Q' = frac{1}{a^2b^2}$$
The area of an ellipse is $pi a b$ so that:
$$text{Area} = pi a b = frac{pi}{sqrt{det Q'}} = frac{picdot |Pcdot t+R|}{sqrt{|det Q|}}
=frac{picdot |-frac 1{13}(4cdot 6 + 4cdot 8)+1|}{sqrt{|6cdot 5-2cdot 2|}}
=frac{43pi}{13sqrt{26}}$$
The general formula is:
$$text{Area}=frac{pi|P^TQ^{-1}P-R|}{sqrt{|det Q|}}$$
answered Jan 14 at 20:43
I like SerenaI like Serena
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$endgroup$
Write the equation as:
$$begin{pmatrix}x & y & 1end{pmatrix}
left(begin{array}{ccc|c} \&Q& &P \ \ hline &P^T& &R end{array}right)
begin{pmatrix}x \ y \ 1end{pmatrix}
=begin{pmatrix}x & y & 1end{pmatrix}
left(begin{array}{cc|c}6 & 2 & 4 \ 2 & 5 & 4 \ hline 4&4&1 end{array}right)
begin{pmatrix}x \ y \ 1end{pmatrix} = 0$$
Set the gradient equal to zero to find the translation (same as solving $Qx=-P$):
$$nabla(6x^2+4xy+5y^2+8x+8y+1)=0 Rightarrow begin{cases}12x+4y+8=0\4x+10y+8=0end{cases} Rightarrow x=-frac 6{13}, y=-frac 8{13}$$
So the translation vector $t=-frac 1{13}begin{pmatrix}6\8end{pmatrix}$.
Divide the equation by $-(Pcdot t + R)$ to normalize it. The resulting $Q'$ is:
$$Q'=-frac{1}{Pcdot t + R}Q$$
Its determinant is the product of the eigenvalues, which are $frac 1{a^2}$ and $frac 1{b^2}$.
So:
$$det Q' = frac{1}{a^2b^2}$$
The area of an ellipse is $pi a b$ so that:
$$text{Area} = pi a b = frac{pi}{sqrt{det Q'}} = frac{picdot |Pcdot t+R|}{sqrt{|det Q|}}
=frac{picdot |-frac 1{13}(4cdot 6 + 4cdot 8)+1|}{sqrt{|6cdot 5-2cdot 2|}}
=frac{43pi}{13sqrt{26}}$$
The general formula is:
$$text{Area}=frac{pi|P^TQ^{-1}P-R|}{sqrt{|det Q|}}$$
answered Jan 14 at 20:43
I like SerenaI like Serena
4,2071722
edited Jan 14 at 20:58
answered Jan 14 at 20:43
I like SerenaI like Serena
4,2071722
answered Jan 14 at 20:43
I like SerenaI like Serena
4,2071722
answered Jan 14 at 20:43
I like SerenaI like Serena
4,2071722
4,2071722
add a comment |
add a comment |
$begingroup$
The simplest way to compute the area is that of using the formulas given here to find semi-axes $a$ and $b$ and then area $=pi ab$.
If you insist on using an integral, you can of course solve for $y$ to find two solutions
$$
y_text{up/down}={1over5}left(
-2x-4pmsqrt{-26x^2-24x+3}
right)
$$
and then compute the area as
$$
int_{x_text{left}}^{x_text{right}}(y_{up}-y_{down}),dx,
$$
where $x_text{left/right}$ are the values of $x$ giving a vanishing square root.
answered Jan 13 at 14:18
AretinoAretino
23.6k21443
$endgroup$
add a comment |
$begingroup$
The simplest way to compute the area is that of using the formulas given here to find semi-axes $a$ and $b$ and then area $=pi ab$.
If you insist on using an integral, you can of course solve for $y$ to find two solutions
$$
y_text{up/down}={1over5}left(
-2x-4pmsqrt{-26x^2-24x+3}
right)
$$
and then compute the area as
$$
int_{x_text{left}}^{x_text{right}}(y_{up}-y_{down}),dx,
$$
where $x_text{left/right}$ are the values of $x$ giving a vanishing square root.
answered Jan 13 at 14:18
AretinoAretino
23.6k21443
$endgroup$
add a comment |
$begingroup$
The simplest way to compute the area is that of using the formulas given here to find semi-axes $a$ and $b$ and then area $=pi ab$.
If you insist on using an integral, you can of course solve for $y$ to find two solutions
$$
y_text{up/down}={1over5}left(
-2x-4pmsqrt{-26x^2-24x+3}
right)
$$
and then compute the area as
$$
int_{x_text{left}}^{x_text{right}}(y_{up}-y_{down}),dx,
$$
where $x_text{left/right}$ are the values of $x$ giving a vanishing square root.
answered Jan 13 at 14:18
AretinoAretino
23.6k21443
$endgroup$
The simplest way to compute the area is that of using the formulas given here to find semi-axes $a$ and $b$ and then area $=pi ab$.
If you insist on using an integral, you can of course solve for $y$ to find two solutions
$$
y_text{up/down}={1over5}left(
-2x-4pmsqrt{-26x^2-24x+3}
right)
$$
and then compute the area as
$$
int_{x_text{left}}^{x_text{right}}(y_{up}-y_{down}),dx,
$$
where $x_text{left/right}$ are the values of $x$ giving a vanishing square root.
answered Jan 13 at 14:18
AretinoAretino
23.6k21443
answered Jan 13 at 14:18
AretinoAretino
23.6k21443
answered Jan 13 at 14:18
AretinoAretino
23.6k21443
answered Jan 13 at 14:18
AretinoAretino
23.6k21443
23.6k21443
add a comment |
add a comment |
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$begingroup$
Hint: Complete the Squares, Shift the centre to the origin and then convert to polar coordinates and integrate
$endgroup$
– DavidG
Jan 13 at 12:32
1
$begingroup$
Following David's comment, if you know how to classify quadratics with matrices and etc., say, you can use that to "make" your ellipse a canonical one (i.e. with center at the origin) and then integrate there.
$endgroup$
– DonAntonio
Jan 13 at 12:34
$begingroup$
@DavidG, when completing the squares, the term $4xy$ disturbs. I have reached to $6(x+frac{2}{3})^2+5(y+frac{4}{5})^2=-4xy+frac{73}{15}$
$endgroup$
– Hussain-Alqatari
Jan 13 at 13:12