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Stereographic projection of ellipsoid

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0

$begingroup$

I am really new in geometry and especially in working with stereographic projection, so excuse me, please, if my question is too easy.

Given is the ellipsoid: $E = left (x,y,z)in mathbbR^3: fracx^2a^2 + fracy^2b^2 +fracz^2c^2 = 1right $.

I have to find two parametrizations with the following points excluded: $Esetminus(a,0,0)$ and $Esetminus(0,0,-a)$.

OK, we know the definition of the stereographic projection of the unit sphere in $mathbbR^3$ with excluding the north pole $(0,0,1)$. It is given by: $(x,y,z)=left ( frac2xx^2+y^2+1, frac2yx^2+y^2+1, fracx^2+y^2-1x^2+y^2+1 right )$.

I know, the problem takes much time to do the calculations, so i would be very glad if someone could give a hint how to do this calculation, because i really don't get it. Thank you very much in advance.

geometry geometric-topology

share|cite|improve this question

edited Mar 15 at 17:06

MarianD

1,5291616

asked Feb 5 '13 at 19:59

LullabyLullaby

917718

$endgroup$

add a comment | 

0

$begingroup$

I am really new in geometry and especially in working with stereographic projection, so excuse me, please, if my question is too easy.

Given is the ellipsoid: $E = left (x,y,z)in mathbbR^3: fracx^2a^2 + fracy^2b^2 +fracz^2c^2 = 1right $.

I have to find two parametrizations with the following points excluded: $Esetminus(a,0,0)$ and $Esetminus(0,0,-a)$.

OK, we know the definition of the stereographic projection of the unit sphere in $mathbbR^3$ with excluding the north pole $(0,0,1)$. It is given by: $(x,y,z)=left ( frac2xx^2+y^2+1, frac2yx^2+y^2+1, fracx^2+y^2-1x^2+y^2+1 right )$.

I know, the problem takes much time to do the calculations, so i would be very glad if someone could give a hint how to do this calculation, because i really don't get it. Thank you very much in advance.

geometry geometric-topology

share|cite|improve this question

edited Mar 15 at 17:06

MarianD

1,5291616

asked Feb 5 '13 at 19:59

LullabyLullaby

917718

$endgroup$

add a comment | 

$begingroup$

I am really new in geometry and especially in working with stereographic projection, so excuse me, please, if my question is too easy.

Given is the ellipsoid: $E = left (x,y,z)in mathbbR^3: fracx^2a^2 + fracy^2b^2 +fracz^2c^2 = 1right $.

I have to find two parametrizations with the following points excluded: $Esetminus(a,0,0)$ and $Esetminus(0,0,-a)$.

OK, we know the definition of the stereographic projection of the unit sphere in $mathbbR^3$ with excluding the north pole $(0,0,1)$. It is given by: $(x,y,z)=left ( frac2xx^2+y^2+1, frac2yx^2+y^2+1, fracx^2+y^2-1x^2+y^2+1 right )$.

I know, the problem takes much time to do the calculations, so i would be very glad if someone could give a hint how to do this calculation, because i really don't get it. Thank you very much in advance.

geometry geometric-topology

share|cite|improve this question

edited Mar 15 at 17:06

MarianD

1,5291616

asked Feb 5 '13 at 19:59

LullabyLullaby

917718

$endgroup$

share|cite|improve this question

edited Mar 15 at 17:06

MarianD

1,5291616

asked Feb 5 '13 at 19:59

LullabyLullaby

917718

share|cite|improve this question

edited Mar 15 at 17:06

MarianD

1,5291616

asked Feb 5 '13 at 19:59

LullabyLullaby

917718

edited Mar 15 at 17:06

MarianD

1,5291616

edited Mar 15 at 17:06

MarianD

1,5291616

asked Feb 5 '13 at 19:59

LullabyLullaby

917718

asked Feb 5 '13 at 19:59

LullabyLullaby

917718

1 Answer
1

active

oldest

votes

0

$begingroup$

A hint for the first one:

Any point $P:=(0,u,v)$ in the $(y,z)$-plane determines a line $g_P:=Pvee A$, where $A:=(a,0,0)$. Intersecting
$$g_P:quad tmapsto bigl((1-t) a,t u,t vbigr)qquad(-infty<t<infty)$$
with the ellipsoid $E$ you get a quadratic equation for $t$ with one obvious solution $t=0$. The other solution leads you to the point $(x_P,y_P,z_P)in E$ stereographically related to $P$. All in all you will obtain a parametric representation
$$(u,v)mapstobigl(x_(u,v),y_(u,v),z_(u,v)bigr)in Esetminus(a,0,0) .$$

share|cite|improve this answer

answered Feb 5 '13 at 20:30

Christian BlatterChristian Blatter

175k8115327

$endgroup$

add a comment | 

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0

$begingroup$

A hint for the first one:

Any point $P:=(0,u,v)$ in the $(y,z)$-plane determines a line $g_P:=Pvee A$, where $A:=(a,0,0)$. Intersecting
$$g_P:quad tmapsto bigl((1-t) a,t u,t vbigr)qquad(-infty<t<infty)$$
with the ellipsoid $E$ you get a quadratic equation for $t$ with one obvious solution $t=0$. The other solution leads you to the point $(x_P,y_P,z_P)in E$ stereographically related to $P$. All in all you will obtain a parametric representation
$$(u,v)mapstobigl(x_(u,v),y_(u,v),z_(u,v)bigr)in Esetminus(a,0,0) .$$

share|cite|improve this answer

answered Feb 5 '13 at 20:30

Christian BlatterChristian Blatter

175k8115327

$endgroup$

add a comment | 

0

$begingroup$

A hint for the first one:

Any point $P:=(0,u,v)$ in the $(y,z)$-plane determines a line $g_P:=Pvee A$, where $A:=(a,0,0)$. Intersecting
$$g_P:quad tmapsto bigl((1-t) a,t u,t vbigr)qquad(-infty<t<infty)$$
with the ellipsoid $E$ you get a quadratic equation for $t$ with one obvious solution $t=0$. The other solution leads you to the point $(x_P,y_P,z_P)in E$ stereographically related to $P$. All in all you will obtain a parametric representation
$$(u,v)mapstobigl(x_(u,v),y_(u,v),z_(u,v)bigr)in Esetminus(a,0,0) .$$

share|cite|improve this answer

answered Feb 5 '13 at 20:30

Christian BlatterChristian Blatter

175k8115327

$endgroup$

add a comment | 

$begingroup$

A hint for the first one:

Any point $P:=(0,u,v)$ in the $(y,z)$-plane determines a line $g_P:=Pvee A$, where $A:=(a,0,0)$. Intersecting
$$g_P:quad tmapsto bigl((1-t) a,t u,t vbigr)qquad(-infty<t<infty)$$
with the ellipsoid $E$ you get a quadratic equation for $t$ with one obvious solution $t=0$. The other solution leads you to the point $(x_P,y_P,z_P)in E$ stereographically related to $P$. All in all you will obtain a parametric representation
$$(u,v)mapstobigl(x_(u,v),y_(u,v),z_(u,v)bigr)in Esetminus(a,0,0) .$$

share|cite|improve this answer

answered Feb 5 '13 at 20:30

Christian BlatterChristian Blatter

175k8115327

$endgroup$

share|cite|improve this answer

answered Feb 5 '13 at 20:30

Christian BlatterChristian Blatter

175k8115327

answered Feb 5 '13 at 20:30

Christian BlatterChristian Blatter

175k8115327

answered Feb 5 '13 at 20:30

Christian BlatterChristian Blatter

175k8115327