In the previous post we established that sections of a right circular
cone are the familiar curves (ellipse, parabola, hyperbola) having the
focus directrix property. Now we will have a look at the more general
case when the cone is not right but oblique. Our approach will be
identical to the one followed by the great Greek geometer Apollonius.
However we will not be developing a systematic theory of conics as
described by Apollonius, but rather focus on the interesting results
which will help us to connect them with the modern definition of conics.
In doing so we will observe that the main tool used by Apollonius is
the similarity of triangles.

Let's have a look at the following axial triangles which are supposed to be perpendicular to the base.

In the above figure we see that line DE intersects the axial triangle in points D and E. In order that the triangles ABC and ADE be similar we must have the corresponding angles equal in both the triangles. Since both the triangles have a common angle A, we only need to make angle D equal to angle B and angle E equal to angle C (or make angle D equal to angle C and angle E equal to angle B). In the case of right circular cone we see that angle B itself is equal to angle C and hence in any case we have only option that the line DE should be parallel to BC. In the language of planes and cones, the cutting plane should be parallel to the base.

However in the case of an oblique cone we have two options. First where the line DE is parallel to BC (so that angle D is equal to angle B) and second when angle D' is equal to angle C. In this case we have two triangles ADE and AE'D' which are both similar to the axial triangle ABC.

In case of the oblique cone, the second plane corresponding to D'E' is said to be subcontrary or antiparallel to the base. We will now establish that for the plane antiparallel to the base the section is a circle.

In the above figure ABC is an axial triangle perpendicular to the base BC of an oblique cone. A plane subcontrary to the base intersects the axial triangle in points H and K, so that triangle AKH is similar to axial triangle ABC. Let P be a point on the section so generated by the intersection of this plane and the cone. We draw a plane parallel to the base which contains point P and suppose that this plane intersects the axial triangle in points D and E. Both the subcontrary plane and the plane parallel to the base meet in a line which intersects axial plane in point M. From the construction it should be clear that the triangles MDH and MKE are similar.

Since both the plane DE and HK are perpendicular to the axial plane, it follows that their intersection PM is perpendicular to the axial plane and hence perpendicular to both line DE and HK. Next it is easy to observe and prove that any line which perpendicular to the axial plane ABC will be bisected by the plane ABC (this happens only when the axial plane itself is perpendicular to the base as is the case here; the reader should try to supply a proof of this; a generalization of this result will be proved later in this post). Since PM is perpedicular to DE therefore it follows that if the PM is produced to make a chord of the circle DPE then it will be bisected by point M and therefore it follows that DE is a diameter of the circle DPE.

Now we can easily see that

$ {PM}^{2} = DM \cdot ME$

and by the similarity of triangles MDH and MKE we have

$ \displaystyle \frac{DM}{HM} = \frac{KM}{EM}\,\,\Rightarrow DM \cdot ME = HM \cdot MK$

and thus we have $ {PM}^{2} = HM \cdot MK$. It follows that the point P lies on a circle with diameter HK. Since P was an arbitrary point on the section generated by subcontrary plane HK, it follows that the section is a circle.

The above result is one of the most suprising and easy to establish in the theory of conics. The surprise element comes because of the non-obviousness of the result as well as the simple proof in contrast of the non-obviousness.

Before we begin further discussion of the conics, let us establish a generalization of a result used in the previous proof. Let us then suppose that ABC is an axial triangle of a cone with vertex A and base BC (so that BC is a diameter of the circular base). It is not necessary to assume that the axial triangle is perpendicular to the base. From a point H on the circumference of the base a perpendicular to the diameter BC is drawn to meet BC in K.

Let generator AQ meet base BC in F (if necessary extend AQ). Through F we draw a perpendicular to BC which meets BC in L and on extending meets the circular base in F'. Clearly FLF' is parallel to HK and therefore parallel to QQ'. It is now clear that Q' also lies on the generator AF'. We join AL and let this line meet QQ' prime in V. By similarity of triangles we can see that QV/FL = Q'V/F'L. Since L is mid point of FF' it follows that QV = Q'V so that QQ' is bisected at V.

In case of circles we know that this line turns out to be a diameter which is perpendicular to the chords in question. But for general conic this line may or may not be perpendicular to the chords it is bisecting. We call such a line a

We now establish the existence of a diameter.

Let a cone be cut by a plane not passing through its vertex A. Suppose that the plane intersects the base of the cone in DE and let BC be the diameter perpendicular to BC. Let BC meet DE in M, so that M is mid point of DE. The plane of section meets the axial plane ABC in line PM where P lies on the section. Let Q be any point on the section apart from P and we draw a line from Q parallel to DE to meet the cone in another point Q'. Clearly by the previous result the chord QQ' gets bisected by the line PM. Thus we see that all the chords parallel to DE are bisected by a single line PM. Thus PM is a diameter. Clearly PM will be perpendicular to the chords it bisects only when the axial plane ABC is perpendicular to the base.

We can summarize the above result as follows:

In the scenario described in above proof, the segment QV is said to be an

Clearly if the cone is right circular then all axial triangles are perpendicular to the base and hence the diameter obtained by intersection with the axial triangle will be perpendicular to the chords it bisects and hence will be an axis of the conic section and clearly the conic section will be symmetrical with respect to the axis.

### Sections of an Oblique Cone: Circle

It is easy to convince ourselves that any plane parallel to the base of an oblique cone will make a circular section and its point of intersection with the axis will be the center of the circle. However it is very surprising that there is another family of parallel planes which intersect the oblique cone in a circular section. Also this is something very unique to the oblique cones and this scenario does not occur for a right circular cone. We have the following theorem by Apollonius:*Let there be an axial plane which is perpendicular to the base of the cone. Let a plane which is perpendicular to this axial plane cut the cone in such a way that the portion of the axial triangle cut by the plane is similar to the whole axial triangle. Then the section of the cone so generated is a circle.*Let's have a look at the following axial triangles which are supposed to be perpendicular to the base.

In the above figure we see that line DE intersects the axial triangle in points D and E. In order that the triangles ABC and ADE be similar we must have the corresponding angles equal in both the triangles. Since both the triangles have a common angle A, we only need to make angle D equal to angle B and angle E equal to angle C (or make angle D equal to angle C and angle E equal to angle B). In the case of right circular cone we see that angle B itself is equal to angle C and hence in any case we have only option that the line DE should be parallel to BC. In the language of planes and cones, the cutting plane should be parallel to the base.

However in the case of an oblique cone we have two options. First where the line DE is parallel to BC (so that angle D is equal to angle B) and second when angle D' is equal to angle C. In this case we have two triangles ADE and AE'D' which are both similar to the axial triangle ABC.

In case of the oblique cone, the second plane corresponding to D'E' is said to be subcontrary or antiparallel to the base. We will now establish that for the plane antiparallel to the base the section is a circle.

Subcontrary Plane

In the above figure ABC is an axial triangle perpendicular to the base BC of an oblique cone. A plane subcontrary to the base intersects the axial triangle in points H and K, so that triangle AKH is similar to axial triangle ABC. Let P be a point on the section so generated by the intersection of this plane and the cone. We draw a plane parallel to the base which contains point P and suppose that this plane intersects the axial triangle in points D and E. Both the subcontrary plane and the plane parallel to the base meet in a line which intersects axial plane in point M. From the construction it should be clear that the triangles MDH and MKE are similar.

Since both the plane DE and HK are perpendicular to the axial plane, it follows that their intersection PM is perpendicular to the axial plane and hence perpendicular to both line DE and HK. Next it is easy to observe and prove that any line which perpendicular to the axial plane ABC will be bisected by the plane ABC (this happens only when the axial plane itself is perpendicular to the base as is the case here; the reader should try to supply a proof of this; a generalization of this result will be proved later in this post). Since PM is perpedicular to DE therefore it follows that if the PM is produced to make a chord of the circle DPE then it will be bisected by point M and therefore it follows that DE is a diameter of the circle DPE.

Now we can easily see that

$ {PM}^{2} = DM \cdot ME$

and by the similarity of triangles MDH and MKE we have

$ \displaystyle \frac{DM}{HM} = \frac{KM}{EM}\,\,\Rightarrow DM \cdot ME = HM \cdot MK$

and thus we have $ {PM}^{2} = HM \cdot MK$. It follows that the point P lies on a circle with diameter HK. Since P was an arbitrary point on the section generated by subcontrary plane HK, it follows that the section is a circle.

The above result is one of the most suprising and easy to establish in the theory of conics. The surprise element comes because of the non-obviousness of the result as well as the simple proof in contrast of the non-obviousness.

Before we begin further discussion of the conics, let us establish a generalization of a result used in the previous proof. Let us then suppose that ABC is an axial triangle of a cone with vertex A and base BC (so that BC is a diameter of the circular base). It is not necessary to assume that the axial triangle is perpendicular to the base. From a point H on the circumference of the base a perpendicular to the diameter BC is drawn to meet BC in K.

*If from a point Q on the cone not lying on axial plane ABC, we draw a line parallel to HK such that it meets the cone in another point Q', then the the line segment QQ' is bisected by the axial plane ABC.*Let generator AQ meet base BC in F (if necessary extend AQ). Through F we draw a perpendicular to BC which meets BC in L and on extending meets the circular base in F'. Clearly FLF' is parallel to HK and therefore parallel to QQ'. It is now clear that Q' also lies on the generator AF'. We join AL and let this line meet QQ' prime in V. By similarity of triangles we can see that QV/FL = Q'V/F'L. Since L is mid point of FF' it follows that QV = Q'V so that QQ' is bisected at V.

### Chords and Diameter of a Conic Section

Next we define some terms. A line segment joining two points on a conic section is called a*chord*of the conic section. We now establish a fundamental property of all conic sections.*A series of parallel chords of a conic section are bisected by a single line*In case of circles we know that this line turns out to be a diameter which is perpendicular to the chords in question. But for general conic this line may or may not be perpendicular to the chords it is bisecting. We call such a line a

*diameter*of the conic section. If a diameter of a conic section is perpendicular to the chords it is bisecting then it is called an*axis*of the conic section. A conic section may have one axis (parabola), two axes (ellipse and hyperbola) or an infinity of axes (circle) and the conic is symmetrical about any of its axes.We now establish the existence of a diameter.

Let a cone be cut by a plane not passing through its vertex A. Suppose that the plane intersects the base of the cone in DE and let BC be the diameter perpendicular to BC. Let BC meet DE in M, so that M is mid point of DE. The plane of section meets the axial plane ABC in line PM where P lies on the section. Let Q be any point on the section apart from P and we draw a line from Q parallel to DE to meet the cone in another point Q'. Clearly by the previous result the chord QQ' gets bisected by the line PM. Thus we see that all the chords parallel to DE are bisected by a single line PM. Thus PM is a diameter. Clearly PM will be perpendicular to the chords it bisects only when the axial plane ABC is perpendicular to the base.

We can summarize the above result as follows:

*If a cone is cut by a plane not passing through the vertex of the cone and which intersects the base of the cone in a straight line perpendicular to the base of an axial triangle, then the intersection of the plane with this axial triangle is a diameter of the conic section generated by the cutting plane.*In the scenario described in above proof, the segment QV is said to be an

*ordinate*on the diameter PM.Clearly if the cone is right circular then all axial triangles are perpendicular to the base and hence the diameter obtained by intersection with the axial triangle will be perpendicular to the chords it bisects and hence will be an axis of the conic section and clearly the conic section will be symmetrical with respect to the axis.

*In the next post we will study about the parabola, ellipse and hyperbola as sections of an oblique cone.***P.S.**: The figures with brownish background are taken from*"Apollonius of Perga: Treatise on Conic Sections"*by T. L. Heath.**Print/PDF Version**
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