After having dealt with the case of circular sections of an oblique cone in previous post,
let's now focus on the conics. We will first treat the case of a
parabola as this is the simplest case after the case of a circular
section.

We have the following property of the parabolas which at once gives the cartesian equation of the parabola:

First of all it is easy to observe that the length PL is a constant which depends only on the cone and the cutting plane. Then the above result says that for the case of a parabola the square of the ordinate on a diameter is proportional to the distance of its foot from the end of the corresponding diameter. It reminds us of the equation $ y^{2} = 4ax$ where $ y$ is the ordinate and $ x$ is the distance of its foot from the vertex of the parabola. The standard equation is for the case when the ordinate is perpendicular to the diameter, but as the above result says a similar relation holds between any ordinate and its corresponding diameter.

Let HK parallel to BC be drawn in the plane of axial triangle ABC such that HK passes through V where QV is an ordinate to the diameter PM. Also we are given that PM is parallel to AC. Clearly the points H, K, Q lie on a circle with HK as diameter and hence

$ \displaystyle {QV}^{2} = HV \cdot VK$

By similarity of triangles we have

$ \displaystyle HV/PV = BC/AC,\,\,\, VK/PA = BC/BA$

so that

$ \displaystyle \frac{HV}{PV}\cdot\frac{VK}{PA} = \frac{{BC}^{2}}{BA \cdot AC} = \frac{PL}{PA}$

$ \displaystyle \Rightarrow HV \cdot VK = PL \cdot PV,\,\,\, \Rightarrow {QV}^{2} = PL \cdot PV$

The line segment PL is called

For an ordinate and its corresponding diameter of a hyperbola the following property holds:

Let HK be a line parallel to BC through V. Then clearly points H, Q, K lie on circle with diameter HK. Hence $ {QV}^{2} = HV \cdot VK$. By similarity of triangles we have

$ \displaystyle \frac{HV}{PV} = \frac{BF}{AF},\,\,\, \frac{VK}{P'V} = \frac{FC}{AF}$

$ \displaystyle \Rightarrow \frac{HV}{PV}\cdot\frac{VK}{P'V} = \frac{BF\cdot FC}{{AF}^{2}}$

$ \displaystyle \Rightarrow \frac{{QV}^{2}}{PV\cdot P'V} = \frac{PL}{PP'}$

$ \displaystyle \Rightarrow {QV}^{2} = PV \cdot\frac{P'V}{PP'}\cdot PL = PV \cdot\frac{VR}{PL}\cdot PL = PV \cdot VR$

Again the length PL is called the

For ellipses we have the following relationship between an ordinate and its diameter:

Let HK be a line parallel to BC through V. Then clearly points H, Q, K lie on circle with diameter HK. Hence $ {QV}^{2} = HV \cdot VK$. By similarity of triangles we have

$ \displaystyle \frac{HV}{PV} = \frac{BF}{AF},\,\,\, \frac{VK}{P'V} = \frac{FC}{AF}$

$ \displaystyle \Rightarrow \frac{HV}{PV}\cdot\frac{VK}{P'V} = \frac{BF\cdot FC}{{AF}^{2}}$

$ \displaystyle \Rightarrow \frac{{QV}^{2}}{PV\cdot P'V} = \frac{PL}{PP'}$

$ \displaystyle \Rightarrow {QV}^{2} = PV \cdot\frac{P'V}{PP'}\cdot PL = PV \cdot\frac{VR}{PL}\cdot PL = PV \cdot VR$

Again the length PL is called the

Thus we have shown the connection of parabola, hyperbola and ellipse with an oblique cone. For further reading of the beautiful theorems by Apollonius I refer the excellent presentation by

### Sections of an Oblique Cone: Parabola

We have so far established in the previous post that if a plane intersects the base of an axial triangle at right angles, then the intersection of this plane and the corresponding axial triangle forms a diameter of the corresponding conic section. If this diameter is parallel to one of the sides of the axial triangle then the section formed is called a*parabola*.We have the following property of the parabolas which at once gives the cartesian equation of the parabola:

*Let the diameter PM of the section of a cone be parallel to one of the sides of the axial triangle (say AC) and let QV be an ordinate to PM. If a line segment PL be drawn perpendicular in PM and in the plane of the section such that $ PL/PA = {BC}^{2}/(BA \cdot AC)$ then $ {QV}^{2} = PL \cdot PV$.*First of all it is easy to observe that the length PL is a constant which depends only on the cone and the cutting plane. Then the above result says that for the case of a parabola the square of the ordinate on a diameter is proportional to the distance of its foot from the end of the corresponding diameter. It reminds us of the equation $ y^{2} = 4ax$ where $ y$ is the ordinate and $ x$ is the distance of its foot from the vertex of the parabola. The standard equation is for the case when the ordinate is perpendicular to the diameter, but as the above result says a similar relation holds between any ordinate and its corresponding diameter.

Parabola

Let HK parallel to BC be drawn in the plane of axial triangle ABC such that HK passes through V where QV is an ordinate to the diameter PM. Also we are given that PM is parallel to AC. Clearly the points H, K, Q lie on a circle with HK as diameter and hence

$ \displaystyle {QV}^{2} = HV \cdot VK$

By similarity of triangles we have

$ \displaystyle HV/PV = BC/AC,\,\,\, VK/PA = BC/BA$

so that

$ \displaystyle \frac{HV}{PV}\cdot\frac{VK}{PA} = \frac{{BC}^{2}}{BA \cdot AC} = \frac{PL}{PA}$

$ \displaystyle \Rightarrow HV \cdot VK = PL \cdot PV,\,\,\, \Rightarrow {QV}^{2} = PL \cdot PV$

The line segment PL is called

*latus rectum*corresponding to the diameter PM.### Sections of an Oblique Cone: Hyperbola

If a diameter of the conic section is not parallel to any side of the corresponding axial triangle, but intersects one side and the other side produced beyond the vertex of the cone, then the section is called a hyperbola.For an ordinate and its corresponding diameter of a hyperbola the following property holds:

*Let PM be a diameter which intersects side AB of axial triangle ABC in P and side AC produced in opposite direction at point P'. Also let AF be a line parallel to PM in the plane of axial triangle and meet the base BC in F. Let PL be perpendicular to diameter PM and in the plane of section such that $ PL/PP' = (BF \cdot FC) / {AF}^{2}$. If QV is an ordinate to diameter PM and VR be drawn parallel to PL and P'L produced meet VR in R, then $ {QV}^{2} = PV \cdot VR$.*Let HK be a line parallel to BC through V. Then clearly points H, Q, K lie on circle with diameter HK. Hence $ {QV}^{2} = HV \cdot VK$. By similarity of triangles we have

$ \displaystyle \frac{HV}{PV} = \frac{BF}{AF},\,\,\, \frac{VK}{P'V} = \frac{FC}{AF}$

$ \displaystyle \Rightarrow \frac{HV}{PV}\cdot\frac{VK}{P'V} = \frac{BF\cdot FC}{{AF}^{2}}$

$ \displaystyle \Rightarrow \frac{{QV}^{2}}{PV\cdot P'V} = \frac{PL}{PP'}$

$ \displaystyle \Rightarrow {QV}^{2} = PV \cdot\frac{P'V}{PP'}\cdot PL = PV \cdot\frac{VR}{PL}\cdot PL = PV \cdot VR$

Again the length PL is called the

*latus rectum*for the diameter PM. If we have $ PL = a, P'P = b, QV = y, PV = x$ then we can easily see that $ y^{2} = ax + (a/b)x^{2}$. This is clearly an equation of the hyperbola as we know from coordinate geometry.### Sections of an Oblique Cone: Ellipse

If a diameter PM of a conic section intersects both sides of an axial triangle and the plane generating the section is neither parallel to the base nor antiparallel to it, then the section is called an ellipse.For ellipses we have the following relationship between an ordinate and its diameter:

*Let a diameter PM of a conic section intersect both sides of the corresponding axial triangle ABC in points P and P'. Let AF be drawn parallel to PP' in the plane of axial triangle such that it meets base BC produced in F. Let PL be a perpendicular to PP' in the plane of the section such that $ PL/PP' = (BF \cdot FC) / {AF}^{2}$. Let QV be an ordinate to diameter PM and let VR parallel to PL meet P'L in R. Then $ {QV}^{2} = PV \cdot VR$.*Let HK be a line parallel to BC through V. Then clearly points H, Q, K lie on circle with diameter HK. Hence $ {QV}^{2} = HV \cdot VK$. By similarity of triangles we have

$ \displaystyle \frac{HV}{PV} = \frac{BF}{AF},\,\,\, \frac{VK}{P'V} = \frac{FC}{AF}$

$ \displaystyle \Rightarrow \frac{HV}{PV}\cdot\frac{VK}{P'V} = \frac{BF\cdot FC}{{AF}^{2}}$

$ \displaystyle \Rightarrow \frac{{QV}^{2}}{PV\cdot P'V} = \frac{PL}{PP'}$

$ \displaystyle \Rightarrow {QV}^{2} = PV \cdot\frac{P'V}{PP'}\cdot PL = PV \cdot\frac{VR}{PL}\cdot PL = PV \cdot VR$

Again the length PL is called the

*latus rectum*for the diameter PM. If we have $ PL = a, P'P = b, QV = y, PV = x$ then we can easily see that $ y^{2} = ax - (a/b)x^{2}$. This is clearly an equation of the ellipse as we know from coordinate geometry.Thus we have shown the connection of parabola, hyperbola and ellipse with an oblique cone. For further reading of the beautiful theorems by Apollonius I refer the excellent presentation by

*"Apollonius of Perga: Treatise on Conic Sections"*by T. L. Heath.**Print/PDF Version**
Exellent explanation and Drawing!

Thank you.

Anonymous

August 27, 2015 at 3:17 PM