黃亦虹 許慶祥

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 curveof E and P is always an ellipse，and in this case a new parametric equation ofis derived.Based on the obtained parametric equation ofand 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 forto 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(shè)E：x2a2+y2b2+z2c2=1為一個(gè)橢球面，P：px+qy+rz=d為一個(gè)平面.利用Householder變換，證明了E和P 相交當(dāng)且僅當(dāng) λ≥d，其中λ=（ap）2+（bq）2+（cr）2.當(dāng) λ>d時(shí)用新的方法證明了橢球面E和平面P的交線 一定是橢圓，并且給出了該橢圓的參數(shù)方程.利用交線的參數(shù)方程，給出了由所圍成的內(nèi)部區(qū)域的面積公式，進(jìn)而給出了橢圓的長(zhǎng)半軸和短半軸的計(jì)算公式.作為應(yīng)用，又給出了交線 成為一個(gè)圓的充要條件.

關(guān)鍵詞：

橢球面; 平面; 參數(shù)方程; Householder變換; 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： 黃亦虹，許慶祥.平面和橢球面相截所得的橢圓的參數(shù)方程及其應(yīng)用 [J].上海師范大學(xué)學(xué)報(bào)（自然科學(xué)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 curveof 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 ofare derived;see Theorem 2.6.As an application，necessary and sufficient conditions are derived under whichis 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 curverepresents 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

Letbe the intersection curve of the ellipsoid E and the plane P given by （1）.Then the area S of the region S bounded bycan 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，whereis 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，whereis 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 ellipsegiven 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 insuch 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

Letbe 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 byis 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.Letbe 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）.Thenis 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 thatis 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.Letbe the intersection curve of the ellipsoid E and the plane P given by （1）.Thenis a circle if and only if p=q=0.

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