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On the ellipsoid and plane intersection equation
Huang Yihong1， Xu Qingxiang2*
（1.College of Sciences，Shanghai Institute of Technology，Shanghai 201418，China;
2.Mathematics and Science College，Shanghai Normal University，Shanghai 200234，China）
Abstract：
Let E：x2a2+y2b2+z2c2=1 be an ellipsoid and P：p x+q y+r z=d be a plane.Based on the Householder transformation，it is shown that the intersection E∩P is nonempty if and only if λ≥d，where λ=（ap）2+（bq）2+（cr）2.When λ>d，this paper provides a new proof that the intersection curve of E and P is always an ellipse，and in this case a new parametric equation of is derived.Based on the obtained parametric equation of and Stokes formula，we derive a formula for the area of the region bounded by ，and compute its semi-major axis and semi-minor axis.As an application，we get necessary and sufficient conditions for to be a circle.
Key words：
ellipsoid; plane; parametric equation; Householder transformation; Stokes formula
CLC number： O 13; O 172Document code： AArticle ID： 1000-5137（2018）01-0024-07
摘要：
設E：x2a2+y2b2+z2c2=1为一个椭球面，P：px+qy+rz=d为一个平面.利用Householder变换，证明了E和P 相交当且仅当 λ≥d，其中λ=（ap）2+（bq）2+（cr）2.当 λ>d时用新的方法证明了椭球面E和平面P的交线 一定是椭圆，并且给出了该椭圆的参数方程.利用交线的参数方程，给出了由所围成的内部区域的面积公式，进而给出了椭圆的长半轴和短半轴的计算公式.作为应用，又给出了交线 成为一个圆的充要条件.
关键词：
椭球面; 平面; 参数方程; Householder变换; Stokes公式
Received date： 2017-01-11
Foundation item： The National Natural Science Foundation of China （11671261）
Biography： Huang Yihong（1965-），female，lecture，reseach area：Advanced mathematics.E-mail：hyh@sit.edu.cn
*Corresponding author： Xu Qingxiang（1967-），male，professor，research area：Functional Analysis.E-mail：qxxu@shnu.edu.cn
引用格式： 黄亦虹，许庆祥.平面和椭球面相截所得的椭圆的参数方程及其应用 [J].上海师范大学学报（自然科学版），2018，47（1）：24-30.
Citation format： Huang Y H，Xu Q X.On the ellipsoid and plane intersection equation [J].Journal of Shanghai Normal University（Natural Sciences），2018，47（1）：24-30.
1Introduction
Throughout this paper，R，R+ and Rm×n are the sets of the real numbers，the positive numbers and the m×n real matrices，respectively.The notation Rn×1 is simplified to Rn.For any A∈Rm×n，its transpose is denoted by AT.Let In be the identity matrix of order n.
Much attention is paid to the very popular topic of the intersection curve of an ellipsoid and a plane[1-3].Yet，little has been done in the literatures on the application of the Householder transformation and the Stokes formula to this topic，which is the concern of this paper. Let E be an ellipsoid and P be a plane defined respectively by
E：x2a2+y2b2+z2c2=1，P：px+qy+rz=d，（1）
where a，b，c∈R+ and p，q，r，d∈R such that p2+q2+r2≠0.
It is known that the intersection curve of E and P is always an ellipse，and much effort has been made in the study of the semi-axes of .Yet，due to the complexity of computation，it is somehow difficult to derive explicit formulas for the semi-axes of .
The key point of this paper is the usage of the Householder transformation to derive a new parametric equation of ，together with the application of the Stokes formula to find the area S of the region bounded by ;see Theorem 2.4 and Corollary 2.5.Another point of this paper is the characterization of the parallel tangent lines of ，which is combined with the obtained formula for S to deal with the semi-axes of .As a result，explicit formulas for the semi-major axis and the semi-minor axis of are derived;see Theorem 2.6.As an application，necessary and sufficient conditions are derived under which is a circle.
2The main results
Let v∈Rn be nonzero.The Householder matrix Hv associated to v is defined by
Hv=In-2vTv· vvT∈Rn×n.（2）
It is known[4] that HvT=Hv and HvTHv=In，i.e.，Hv is an orthogonal matrix.Due to the following property， the Householder matrix is of special usefulness.
Lemma 2.1
Let x，y∈Rn be such that x≠y and xTx=yTy.Then
Hv（x）=y，where v=x-y.（3）
Theorem 2.2
Let E and P be given by （1）.Then =E∩P≠ if and only if λ≥d，where λ is defined by
λ=（ap）2+（bq）2+（cr）2.（4）
Proof
Let λ be defined by （4）.Firstly，we consider the case that p2+q2>0. Let w1=（ap，bq，cr）T and w2=（0，0，λ）T.Then clearly，w1≠w2 and w1Tw1=w2Tw2，so by Lemma 2.1 we have
Hvw1=w2，where v=w1-w2.（5）
Let
x
y
z=
a
b
c
Hv
1a
1b
1c
x1
y1
z1
.（6）
Then by （1），（5） and （6），we have
1=xa，yb，zcxa，yb，zcT=x1a，y1b，z1cHvTHvx1a，y1b，z1cT
=x1a，y1b，z1cx1a，y1b，z1cT=x12a2+y12b2+z12c2，（7）
d=（p，q，r）（x，y，z）T=（ap，bq，cr）Hvx1a，y1b，z1cT
=w1THvx1a，y1b，z1cT=w2Tx1a，y1b，z1cT=z1λc.
（8）
It follows from （7） and （8） that
x12a2+y12b2=1-d2λ2.（9）
This means E∩P is nonempty if and only if λ≥d，where λ is defined by （4）.
Secondly，we consider the case that p=q=0.In this case，we have r≠0.It follows directly from （1） that x2a2+y2b2=1-d2λ2，
thus the conclusion also holds.
The following result is well-known，yet its proof presented below is somehow new.
Theorem 2.3
Let E，P and λ be given by （1） and （4） respectively such that λ>d. Then the intersection curve =E∩P is always an ellipse.
Proof
It needs only to consider the case that p2+q2>0.Let w3=（p，q，r）T，w4=（0，0，p2+q2+r2）T，v1=w3-w4 and Hv1 be the Householder matrix defined by （2） which satisfies Hv1w3=w4.Let
（x，y，z）T=Hv1（x1，y1，z1）T.（10）
Then
d=w3T（x，y，x）T=w3T Hv1（x1，y1，z1）T=w4T（x1，y1，z1）T=p2+q2+r2z1.
Therefore，
z1=dp2+q2+r2=defk.（11）
It follows from （1），（10） and （11） that
1=（x，y，z）
1a2
1b2
1c2
x
y
z
=（x1，y1，k）A
x1
y1
k
，（12）
where A=Hv1T1a2
1b2
1c2 Hv1 is positive definite.Let A be partitioned as A= A1〖|〗A2 〖-〗
A2T〖|〗a3 ，where A1∈R2×2 is positive definite since A is.Then from （12） we get
（x1，y1）A1x1
y1 +λ1 x1+λ2 y1+λ3=0 for some λi∈R，i=1，2，3，
which represents an ellipse in x1y1- plane since A1 is positive definite.This observation together with （11） yields the fact that in the x1y1z1- coordinate system，the equation of the intersection curve represents an ellipse.The conclusion then follows from （10） since Hv1 is an orthogonal matrix.
Theorem 2.4
Let E，P and λ be given by （1） and （4） such that λ>d.Then a parametric equation of the intersection curve =E∩P can be given for t∈[0，2π] as follows：
x（t）=aλ2-d2λ2（cr-λ）[（λ（cr-λ）+（ap）2）cos（t）+abpqsint]+a2pdλ2，
y（t）=bλ2-d2λ2（cr-λ）[abpqcost+（λ（cr-λ）+（bq）2）sint]+b2qdλ2，
z（t）=cλ2-d2λ2[apcost+bqsint]+c2rdλ2.
（13）
Proof
We only consider the case that p2+q2>0.By （2） and （5） we obtain
Hv=
1+（ap）2λ（cr-λ）abpqλ（cr-λ）apλ
abpqλ（cr-λ）1+（bq）2λ（cr-λ）bqλ
apλbqλcrλ
.（14）
Furthermore，by （8） and （9） we get
x1=a1-d2λ2cost，
y1=b1-d2λ2sint，t∈[0，2π]，
z1=cdλ.
（15）
Eq.（13） then follows from （6），（14） and （15）.
An application of Theorem 2.4 is as follows.
Corollary 2.5
Let be the intersection curve of the ellipsoid E and the plane P given by （1）.Then the area S of the region S bounded by can be formulated by
S=abcp2+q2+r2（λ2-d2）πλ3，（16）
where λ is defined by （4）. Proof
Let （cosα，cosβ，cosγ） denote the unit normal vector of the plane P，where
cosα=pp2+q2+r2，cosβ=qp2+q2+r2，cosγ=rp2+q2+r2.（17）
We may use the Stokes formula to get
S=± ∫ zcosβdx+xcosγdy+ycosαdz，（18）
where ± is chosen to ensure that the right side of （18） is non-negative.Note that
∫2π0sintdt=∫2π0costdt=∫2π0sintcostdt=0，（19）
∫2π0sin2tdt=∫2π0cos2tdt=π.（20）
Therefore，by （13），（4），（19） and （20） we obtain
∫ zdx=-abcq（λ2-d2）πλ3，（21）
∫ xdy=-abcr（λ2-d2）πλ3，（22）
∫ ydz=-abcp（λ2-d2）πλ3.（23）
Formula （16） then follows from （17）-（18） and （21）-（23）.
Consider the calculation of I=∫ x2 ds，where is the intersection curve of the sphere x2+y2+z2=R2 （R>0） and the plane x+y+z=0.In view of the symmetry，a solution can be carried out simply as
I=13 ∫ （x2+y2+z2） ds=13 ∫ R2 ds=13 R2· 2π R=2π3R3.
Obviously，the method employed above only works for the symmetric case.As shown by Example 2.1 below，the parametric equation （13） is a useful tool to deal with the non-symmetric case.
Example 2.1
Evaluate I=∫ x2 ds，where is the intersection curve of the sphere x2+y2+z2=R2 （R>0） and the plane px+qy+rz=d.
Solution
We follow the notations as in the proof of Theorem 2.2.Since a=b=c=R，Eq.（6） turns out to be
（x，y，z）T=Hv （x1，y1，z1）T，
which is combined with （15） to get
ds=（x′（t），y′（t），z′（t））（x′（t），y′（t），z′（t））Tdt
=（x′1（t），y′1（t），z′1（t））HvT Hv（x′1（t），y′1（t），z′1（t））Tdt
=（x′1（t），y′1（t），z′1（t））g（x′1（t），y′1（t），z′1（t））Tdt
=（x′1（t））2+（y′1（t））2dt=R1-d2λ2dt.
In view of the first equation of （13） and （19）-（20），we have
I=∫ x2 ds=μR1-d2λ2，where λ=Rp2+q2+r2，（24）
and μ is given by
μ=R2（λ2-d2）πλ4（Rr-λ）2[（λ（Rr-λ）+（Rp）2）2+（R2pq）2]+2π（R2pd）2λ4.（25）
Note that
（λ（Rr-λ）+（Rp）2）2+（R2pq）2
=λ2（Rr-λ）2+2R2p2λ（Rr-λ）+R2p2· R2（p2+q2）
=λ2（Rr-λ）2+2R2p2λ（Rr-λ）+R2p2（λ2-R2r2）
=（Rr-λ）[λ2（Rr-λ）+2R2p2λ-R2p2（λ+Rr）]
=（Rr-λ）[R2（p2+q2+r2）（Rr-λ）+2R2p2λ-R2p2（λ+Rr）]
=（Rr-λ）2R2（q2+r2），（26）
so we may combine （24）-（26） to conclude that
I=∫ x2 ds=π R5[（λ2-d2）（q2+r2）+2p2d2]λ2-d2λ5，
where λ is given by （24）.
Now，we turn to study the semi-axes of the ellipse given by （13）.Let P（t）=（x（t），y（t），z（t）） be a point in .Then we have
x′（t）=aλ2-d2λ2（cr-λ）[（λ（cr-λ）+（ap）2）（-sint）+abpqcost]，
y′（t）=bλ2-d2λ2（cr-λ）[abpq（-sint）+（λ（cr-λ）+（bq）2）cost]， z′（t）=cλ2-d2λ2[ap（-sint）+bqcost]，
where λ is given by （4）.Suppose that P（t1） and P（t2） are two different points in such that the tangent lines at these two points are parallel，then there exists a constant μ such that x′（t2）=μx′（t1），y′（t2）=μy′（t1） and z′（t2）=μ z′（t1）;or more precisely，
（λ（cr-λ）+（ap）2）（-sint2）+abpqcost2
=μ[（λ（cr-λ）+（ap）2）（-sint1）+abpqcost1]，（27）
abpq（-sint2）+（λ（cr-λ）+（bq）2）cost2
=μ[abpq（-sint1）+（λ（cr-λ）+（bq）2）cost1]，（28）
ap（-sint2）+bqcost2=μ[ap（-sint1）+bqcost1].（29）
It follows from （27） and （29），（28） and （29） that sint2=μsint1 and cost2=μcost1.Therefore，
1=sin2t2+cos2t2=μ2 （sin2t1+cos2t1）=μ2，
hence μ=-1 since P（t1）≠P（t2），and thus P（t2）=P（t1+π）.The observation above indicates that
42max=max{f（t）|t∈[0，2π]}，42min=min{f（t）|t∈[0，2π]}，（30）
where max，min denote the semi-major axis and the semi-minor axis of ，respectively，and
f（t）=[x（t+π）-x（t）]2+[y（t+π）-y（t）]2+[z（t+π）-z（t）]2
= 4（λ2-d2）λ4 g（t），（31）
where g（t） is given by
g（t）=a2（cr-λ）2[（λ（cr-λ）+（ap）2）cos（t）+abpqsin（t）]2
+b2（cr-λ）2[abpqcos（t）+（λ（cr-λ）+（bq）2）sin（t）]2
+c2[apcos（t）+bqsin（t）]2=A+Bcos（2t）+Csin（2t）.（32）
as cos2t=1+cos（2t）2，sin2t=1-cos（2t）2 and sintcost=sin（2t）2，where
A=a22（cr-λ）2[（λ（cr-λ）+（ap）2）2+（abpq）2]
+b22（cr-λ）2[（abpq）2+（λ（cr-λ）+（bq）2）2]+c22[（ap）2+（bq）2].（33）
By （4） we have
[（λ（cr-λ）+（ap）2）2+（abpq）2]=λ2（cr-λ）2+2（ap）2λ（cr-λ）+a2p2[（ap）2+（bq）2]
=λ2（cr-λ）2+2（ap）2λ（cr-λ）+a2p2[λ2-（cr）2]=（cr-λ）2[λ2-（ap）2].（34）
Similarly，we have
[（abpq）2+（λ（cr-λ）+（bq）2）2]=（cr-λ）2[λ2-（bq）2].（35）
We may combine （4） with （33）-（35） to conclude that
A=12[（ap）2（b2+c2）+（bq）2（c2+a2）+（cr）2（a2+b2）].（36）
Theorem 2.6
Let be the intersection curve of the ellipsoid E and the plane P given by （1）， and max and min be the semi-major axis and the semi-minor axis of .Then
max=λ2-d2λ2A+A2-λ2 （abc）2（p2+q2+r2），
min=λ2-d2λ2A-A2-λ2 （abc）2（p2+q2+r2），
where λ and A are defined by （4） and （36）.
Proof
It follows from （30）-（32） that
max=λ2-d2λ2A+B2+C2，（37）
min=λ2-d2λ2A-B2+C2，（38）
which means that the area S of the region S bounded by is equal to
π maxmin=πλ2-d2λ4A2-（B2+C2）.
The above equation together with （16） yields
B2+C2=A2-λ2 （abc）2（p2+q2+r2）.
The conclusion then follows by substituting the above expression for B2+C2 into （37） and （38）. A direct application of the preceding theorem is as follows.
Corollary 2.7
Suppose that a>b>c>0.Let be the intersection curve of the ellipsoid E and the plane P given by （1），and n→=（cosα，cosβ，cosγ） be the unit normal vector of P with cosα，cosβ and cosγ given by （17）.Then is a circle if and only if either n→‖n1 or n→‖n2，where
n1=1b2-1a2，0，1c2-1b2，n2=1b2-1a2，0，-1c2-1b2.
Proof
Let λ and A be defined by （4） and （36）.By direct computation，we have
θ=def4A2-4λ2 （abc）2（p2+q2+r2）=[（cr）2（a2-b2）-（ap）2（b2-c2）]2+（bq）4（a2-c2）2
+2（abpq）2（a2-c2）（b2-c2）+2（bcqr）2（a2-c2）（a2-b2）.
Since a>b>c，by Theorem 2.6 we know that is a circle if and only if θ=0.Equivalently， is a circle if and only if
q=0 and cra2-b2=± apb2-c2n→‖ n1 or n→‖ n2.
The result stated below follows immediately from the proof of Corollary 2.7.
Corollary 2.8
Suppose that a，b，c∈R+ such that a=b≠c.Let be the intersection curve of the ellipsoid E and the plane P given by （1）.Then is a circle if and only if p=q=0.
References：
[1]Abramson N，Boman J，Bonnevier B.Plane intersections of rotational elliposids [J].American Mathematical Monthly，2005，113：336-339.
[2]Ferguson C C.Intersections of ellipsoids and planes of arbitrary orientation and position [J].Mathematical Geology，1979，11：329-336.
[3]Klein P P.On the ellipsoid and plane intersection equation [J].Applied Mathematics，2012，11：1634-1640.
[4]Leon S J.Linear Algebra with Applications （Eighth Edition） [M].Beijing：Pearson Education Asia Limited and China Machine Press，2011.
Huang Yihong1， Xu Qingxiang2*
（1.College of Sciences，Shanghai Institute of Technology，Shanghai 201418，China;
2.Mathematics and Science College，Shanghai Normal University，Shanghai 200234，China）
Abstract：
Let E：x2a2+y2b2+z2c2=1 be an ellipsoid and P：p x+q y+r z=d be a plane.Based on the Householder transformation，it is shown that the intersection E∩P is nonempty if and only if λ≥d，where λ=（ap）2+（bq）2+（cr）2.When λ>d，this paper provides a new proof that the intersection curve of E and P is always an ellipse，and in this case a new parametric equation of is derived.Based on the obtained parametric equation of and Stokes formula，we derive a formula for the area of the region bounded by ，and compute its semi-major axis and semi-minor axis.As an application，we get necessary and sufficient conditions for to be a circle.
Key words：
ellipsoid; plane; parametric equation; Householder transformation; Stokes formula
CLC number： O 13; O 172Document code： AArticle ID： 1000-5137（2018）01-0024-07
摘要：
設E：x2a2+y2b2+z2c2=1为一个椭球面，P：px+qy+rz=d为一个平面.利用Householder变换，证明了E和P 相交当且仅当 λ≥d，其中λ=（ap）2+（bq）2+（cr）2.当 λ>d时用新的方法证明了椭球面E和平面P的交线 一定是椭圆，并且给出了该椭圆的参数方程.利用交线的参数方程，给出了由所围成的内部区域的面积公式，进而给出了椭圆的长半轴和短半轴的计算公式.作为应用，又给出了交线 成为一个圆的充要条件.
关键词：
椭球面; 平面; 参数方程; Householder变换; Stokes公式
Received date： 2017-01-11
Foundation item： The National Natural Science Foundation of China （11671261）
Biography： Huang Yihong（1965-），female，lecture，reseach area：Advanced mathematics.E-mail：hyh@sit.edu.cn
*Corresponding author： Xu Qingxiang（1967-），male，professor，research area：Functional Analysis.E-mail：qxxu@shnu.edu.cn
引用格式： 黄亦虹，许庆祥.平面和椭球面相截所得的椭圆的参数方程及其应用 [J].上海师范大学学报（自然科学版），2018，47（1）：24-30.
Citation format： Huang Y H，Xu Q X.On the ellipsoid and plane intersection equation [J].Journal of Shanghai Normal University（Natural Sciences），2018，47（1）：24-30.
1Introduction
Throughout this paper，R，R+ and Rm×n are the sets of the real numbers，the positive numbers and the m×n real matrices，respectively.The notation Rn×1 is simplified to Rn.For any A∈Rm×n，its transpose is denoted by AT.Let In be the identity matrix of order n.
Much attention is paid to the very popular topic of the intersection curve of an ellipsoid and a plane[1-3].Yet，little has been done in the literatures on the application of the Householder transformation and the Stokes formula to this topic，which is the concern of this paper. Let E be an ellipsoid and P be a plane defined respectively by
E：x2a2+y2b2+z2c2=1，P：px+qy+rz=d，（1）
where a，b，c∈R+ and p，q，r，d∈R such that p2+q2+r2≠0.
It is known that the intersection curve of E and P is always an ellipse，and much effort has been made in the study of the semi-axes of .Yet，due to the complexity of computation，it is somehow difficult to derive explicit formulas for the semi-axes of .
The key point of this paper is the usage of the Householder transformation to derive a new parametric equation of ，together with the application of the Stokes formula to find the area S of the region bounded by ;see Theorem 2.4 and Corollary 2.5.Another point of this paper is the characterization of the parallel tangent lines of ，which is combined with the obtained formula for S to deal with the semi-axes of .As a result，explicit formulas for the semi-major axis and the semi-minor axis of are derived;see Theorem 2.6.As an application，necessary and sufficient conditions are derived under which is a circle.
2The main results
Let v∈Rn be nonzero.The Householder matrix Hv associated to v is defined by
Hv=In-2vTv· vvT∈Rn×n.（2）
It is known[4] that HvT=Hv and HvTHv=In，i.e.，Hv is an orthogonal matrix.Due to the following property， the Householder matrix is of special usefulness.
Lemma 2.1
Let x，y∈Rn be such that x≠y and xTx=yTy.Then
Hv（x）=y，where v=x-y.（3）
Theorem 2.2
Let E and P be given by （1）.Then =E∩P≠ if and only if λ≥d，where λ is defined by
λ=（ap）2+（bq）2+（cr）2.（4）
Proof
Let λ be defined by （4）.Firstly，we consider the case that p2+q2>0. Let w1=（ap，bq，cr）T and w2=（0，0，λ）T.Then clearly，w1≠w2 and w1Tw1=w2Tw2，so by Lemma 2.1 we have
Hvw1=w2，where v=w1-w2.（5）
Let
x
y
z=
a
b
c
Hv
1a
1b
1c
x1
y1
z1
.（6）
Then by （1），（5） and （6），we have
1=xa，yb，zcxa，yb，zcT=x1a，y1b，z1cHvTHvx1a，y1b，z1cT
=x1a，y1b，z1cx1a，y1b，z1cT=x12a2+y12b2+z12c2，（7）
d=（p，q，r）（x，y，z）T=（ap，bq，cr）Hvx1a，y1b，z1cT
=w1THvx1a，y1b，z1cT=w2Tx1a，y1b，z1cT=z1λc.
（8）
It follows from （7） and （8） that
x12a2+y12b2=1-d2λ2.（9）
This means E∩P is nonempty if and only if λ≥d，where λ is defined by （4）.
Secondly，we consider the case that p=q=0.In this case，we have r≠0.It follows directly from （1） that x2a2+y2b2=1-d2λ2，
thus the conclusion also holds.
The following result is well-known，yet its proof presented below is somehow new.
Theorem 2.3
Let E，P and λ be given by （1） and （4） respectively such that λ>d. Then the intersection curve =E∩P is always an ellipse.
Proof
It needs only to consider the case that p2+q2>0.Let w3=（p，q，r）T，w4=（0，0，p2+q2+r2）T，v1=w3-w4 and Hv1 be the Householder matrix defined by （2） which satisfies Hv1w3=w4.Let
（x，y，z）T=Hv1（x1，y1，z1）T.（10）
Then
d=w3T（x，y，x）T=w3T Hv1（x1，y1，z1）T=w4T（x1，y1，z1）T=p2+q2+r2z1.
Therefore，
z1=dp2+q2+r2=defk.（11）
It follows from （1），（10） and （11） that
1=（x，y，z）
1a2
1b2
1c2
x
y
z
=（x1，y1，k）A
x1
y1
k
，（12）
where A=Hv1T1a2
1b2
1c2 Hv1 is positive definite.Let A be partitioned as A= A1〖|〗A2 〖-〗
A2T〖|〗a3 ，where A1∈R2×2 is positive definite since A is.Then from （12） we get
（x1，y1）A1x1
y1 +λ1 x1+λ2 y1+λ3=0 for some λi∈R，i=1，2，3，
which represents an ellipse in x1y1- plane since A1 is positive definite.This observation together with （11） yields the fact that in the x1y1z1- coordinate system，the equation of the intersection curve represents an ellipse.The conclusion then follows from （10） since Hv1 is an orthogonal matrix.
Theorem 2.4
Let E，P and λ be given by （1） and （4） such that λ>d.Then a parametric equation of the intersection curve =E∩P can be given for t∈[0，2π] as follows：
x（t）=aλ2-d2λ2（cr-λ）[（λ（cr-λ）+（ap）2）cos（t）+abpqsint]+a2pdλ2，
y（t）=bλ2-d2λ2（cr-λ）[abpqcost+（λ（cr-λ）+（bq）2）sint]+b2qdλ2，
z（t）=cλ2-d2λ2[apcost+bqsint]+c2rdλ2.
（13）
Proof
We only consider the case that p2+q2>0.By （2） and （5） we obtain
Hv=
1+（ap）2λ（cr-λ）abpqλ（cr-λ）apλ
abpqλ（cr-λ）1+（bq）2λ（cr-λ）bqλ
apλbqλcrλ
.（14）
Furthermore，by （8） and （9） we get
x1=a1-d2λ2cost，
y1=b1-d2λ2sint，t∈[0，2π]，
z1=cdλ.
（15）
Eq.（13） then follows from （6），（14） and （15）.
An application of Theorem 2.4 is as follows.
Corollary 2.5
Let be the intersection curve of the ellipsoid E and the plane P given by （1）.Then the area S of the region S bounded by can be formulated by
S=abcp2+q2+r2（λ2-d2）πλ3，（16）
where λ is defined by （4）. Proof
Let （cosα，cosβ，cosγ） denote the unit normal vector of the plane P，where
cosα=pp2+q2+r2，cosβ=qp2+q2+r2，cosγ=rp2+q2+r2.（17）
We may use the Stokes formula to get
S=± ∫ zcosβdx+xcosγdy+ycosαdz，（18）
where ± is chosen to ensure that the right side of （18） is non-negative.Note that
∫2π0sintdt=∫2π0costdt=∫2π0sintcostdt=0，（19）
∫2π0sin2tdt=∫2π0cos2tdt=π.（20）
Therefore，by （13），（4），（19） and （20） we obtain
∫ zdx=-abcq（λ2-d2）πλ3，（21）
∫ xdy=-abcr（λ2-d2）πλ3，（22）
∫ ydz=-abcp（λ2-d2）πλ3.（23）
Formula （16） then follows from （17）-（18） and （21）-（23）.
Consider the calculation of I=∫ x2 ds，where is the intersection curve of the sphere x2+y2+z2=R2 （R>0） and the plane x+y+z=0.In view of the symmetry，a solution can be carried out simply as
I=13 ∫ （x2+y2+z2） ds=13 ∫ R2 ds=13 R2· 2π R=2π3R3.
Obviously，the method employed above only works for the symmetric case.As shown by Example 2.1 below，the parametric equation （13） is a useful tool to deal with the non-symmetric case.
Example 2.1
Evaluate I=∫ x2 ds，where is the intersection curve of the sphere x2+y2+z2=R2 （R>0） and the plane px+qy+rz=d.
Solution
We follow the notations as in the proof of Theorem 2.2.Since a=b=c=R，Eq.（6） turns out to be
（x，y，z）T=Hv （x1，y1，z1）T，
which is combined with （15） to get
ds=（x′（t），y′（t），z′（t））（x′（t），y′（t），z′（t））Tdt
=（x′1（t），y′1（t），z′1（t））HvT Hv（x′1（t），y′1（t），z′1（t））Tdt
=（x′1（t），y′1（t），z′1（t））g（x′1（t），y′1（t），z′1（t））Tdt
=（x′1（t））2+（y′1（t））2dt=R1-d2λ2dt.
In view of the first equation of （13） and （19）-（20），we have
I=∫ x2 ds=μR1-d2λ2，where λ=Rp2+q2+r2，（24）
and μ is given by
μ=R2（λ2-d2）πλ4（Rr-λ）2[（λ（Rr-λ）+（Rp）2）2+（R2pq）2]+2π（R2pd）2λ4.（25）
Note that
（λ（Rr-λ）+（Rp）2）2+（R2pq）2
=λ2（Rr-λ）2+2R2p2λ（Rr-λ）+R2p2· R2（p2+q2）
=λ2（Rr-λ）2+2R2p2λ（Rr-λ）+R2p2（λ2-R2r2）
=（Rr-λ）[λ2（Rr-λ）+2R2p2λ-R2p2（λ+Rr）]
=（Rr-λ）[R2（p2+q2+r2）（Rr-λ）+2R2p2λ-R2p2（λ+Rr）]
=（Rr-λ）2R2（q2+r2），（26）
so we may combine （24）-（26） to conclude that
I=∫ x2 ds=π R5[（λ2-d2）（q2+r2）+2p2d2]λ2-d2λ5，
where λ is given by （24）.
Now，we turn to study the semi-axes of the ellipse given by （13）.Let P（t）=（x（t），y（t），z（t）） be a point in .Then we have
x′（t）=aλ2-d2λ2（cr-λ）[（λ（cr-λ）+（ap）2）（-sint）+abpqcost]，
y′（t）=bλ2-d2λ2（cr-λ）[abpq（-sint）+（λ（cr-λ）+（bq）2）cost]， z′（t）=cλ2-d2λ2[ap（-sint）+bqcost]，
where λ is given by （4）.Suppose that P（t1） and P（t2） are two different points in such that the tangent lines at these two points are parallel，then there exists a constant μ such that x′（t2）=μx′（t1），y′（t2）=μy′（t1） and z′（t2）=μ z′（t1）;or more precisely，
（λ（cr-λ）+（ap）2）（-sint2）+abpqcost2
=μ[（λ（cr-λ）+（ap）2）（-sint1）+abpqcost1]，（27）
abpq（-sint2）+（λ（cr-λ）+（bq）2）cost2
=μ[abpq（-sint1）+（λ（cr-λ）+（bq）2）cost1]，（28）
ap（-sint2）+bqcost2=μ[ap（-sint1）+bqcost1].（29）
It follows from （27） and （29），（28） and （29） that sint2=μsint1 and cost2=μcost1.Therefore，
1=sin2t2+cos2t2=μ2 （sin2t1+cos2t1）=μ2，
hence μ=-1 since P（t1）≠P（t2），and thus P（t2）=P（t1+π）.The observation above indicates that
42max=max{f（t）|t∈[0，2π]}，42min=min{f（t）|t∈[0，2π]}，（30）
where max，min denote the semi-major axis and the semi-minor axis of ，respectively，and
f（t）=[x（t+π）-x（t）]2+[y（t+π）-y（t）]2+[z（t+π）-z（t）]2
= 4（λ2-d2）λ4 g（t），（31）
where g（t） is given by
g（t）=a2（cr-λ）2[（λ（cr-λ）+（ap）2）cos（t）+abpqsin（t）]2
+b2（cr-λ）2[abpqcos（t）+（λ（cr-λ）+（bq）2）sin（t）]2
+c2[apcos（t）+bqsin（t）]2=A+Bcos（2t）+Csin（2t）.（32）
as cos2t=1+cos（2t）2，sin2t=1-cos（2t）2 and sintcost=sin（2t）2，where
A=a22（cr-λ）2[（λ（cr-λ）+（ap）2）2+（abpq）2]
+b22（cr-λ）2[（abpq）2+（λ（cr-λ）+（bq）2）2]+c22[（ap）2+（bq）2].（33）
By （4） we have
[（λ（cr-λ）+（ap）2）2+（abpq）2]=λ2（cr-λ）2+2（ap）2λ（cr-λ）+a2p2[（ap）2+（bq）2]
=λ2（cr-λ）2+2（ap）2λ（cr-λ）+a2p2[λ2-（cr）2]=（cr-λ）2[λ2-（ap）2].（34）
Similarly，we have
[（abpq）2+（λ（cr-λ）+（bq）2）2]=（cr-λ）2[λ2-（bq）2].（35）
We may combine （4） with （33）-（35） to conclude that
A=12[（ap）2（b2+c2）+（bq）2（c2+a2）+（cr）2（a2+b2）].（36）
Theorem 2.6
Let be the intersection curve of the ellipsoid E and the plane P given by （1）， and max and min be the semi-major axis and the semi-minor axis of .Then
max=λ2-d2λ2A+A2-λ2 （abc）2（p2+q2+r2），
min=λ2-d2λ2A-A2-λ2 （abc）2（p2+q2+r2），
where λ and A are defined by （4） and （36）.
Proof
It follows from （30）-（32） that
max=λ2-d2λ2A+B2+C2，（37）
min=λ2-d2λ2A-B2+C2，（38）
which means that the area S of the region S bounded by is equal to
π maxmin=πλ2-d2λ4A2-（B2+C2）.
The above equation together with （16） yields
B2+C2=A2-λ2 （abc）2（p2+q2+r2）.
The conclusion then follows by substituting the above expression for B2+C2 into （37） and （38）. A direct application of the preceding theorem is as follows.
Corollary 2.7
Suppose that a>b>c>0.Let be the intersection curve of the ellipsoid E and the plane P given by （1），and n→=（cosα，cosβ，cosγ） be the unit normal vector of P with cosα，cosβ and cosγ given by （17）.Then is a circle if and only if either n→‖n1 or n→‖n2，where
n1=1b2-1a2，0，1c2-1b2，n2=1b2-1a2，0，-1c2-1b2.
Proof
Let λ and A be defined by （4） and （36）.By direct computation，we have
θ=def4A2-4λ2 （abc）2（p2+q2+r2）=[（cr）2（a2-b2）-（ap）2（b2-c2）]2+（bq）4（a2-c2）2
+2（abpq）2（a2-c2）（b2-c2）+2（bcqr）2（a2-c2）（a2-b2）.
Since a>b>c，by Theorem 2.6 we know that is a circle if and only if θ=0.Equivalently， is a circle if and only if
q=0 and cra2-b2=± apb2-c2n→‖ n1 or n→‖ n2.
The result stated below follows immediately from the proof of Corollary 2.7.
Corollary 2.8
Suppose that a，b，c∈R+ such that a=b≠c.Let be the intersection curve of the ellipsoid E and the plane P given by （1）.Then is a circle if and only if p=q=0.
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