Is ellipse really a section of cone
- 1. to show that an ellipse is actually a section of a cone: The purpose of this article is to show that conic sections are really intersections of a cone and a plane not through the vertex of the cone; to satisfy the curiosity of the student. Moreover, there is a whole branch of geometry, called projective geometry, in which it is very simply shown that conics are projections of a circle on a plane and general theorems of circles can be transported to conic sections through principles of projective geometry. Let a line segment
- 2. WS = t revolve about a fixed st.line WG always making a constant angle with it at the point W and generate a right circular cone WSK, where K is a point on the cone such that SK is the diameter of the circle obtained by intersection of the cone with a plane passing through S ( a sort of base of the cone) and SK is perpendicular to the axis WG at the point D. Now D is the center of this circle and its radius is given by SD = DK =WS sin = t sin ……………………………….(a) (Note that the triangles WSD and WDK are congruent to each other having a common side WD, a right angle in each and the angle in each of them. Thus DS = DK, and thus the intersection of a plane with a cone perpendicular to its axis happens to be a circle).
- 3. Now let a plane perpendicular to the plane of WSDK pass through the point S making an angle with SK and P be any point on it on its intersection with the cone’s curved surface and let ST be the line of intersection of this plane with the plane of WSDK , meeting the cone at T. Let PST is the curve of intersection with the cone and the plane makes an angle with SK means that ST makes an angle with SK i.e., KST = . Drop a
- 4. perpendicular PQ on ST, meeting the curve of intersection at R. Set up a rectangular Cartesian coordinate system with origin (0, 0) at S, the x-axis along ST and the y- axis along a direction parallel to PQ ( SY in the figure). If (x, y) be the coordinates of the point P , SQ = x and PQ = y………………………………(b) A relationship between x and y shall represent the equation to the curve of intersection SRTP which is our goal; as P is any point on the curve. For that we take a plane containing PR and perpendicular to the axis of the cone WG which intersects the cone in a circle PURV,(as explained in last paragraph) with its center at C on WG and let its radius be UC = PC = CV = CR = r……………………….(c)
- 5. where U and V are points on the cone and UV || SK. We can find the relationship between x and y , by relating them easily to UQ and PC = r. We immediately note that SQU = QSK = ……………………………(d) as SK || UV. To find x, let us concentrate on the triangle SUQ and
- 6. resort to the law of cosines or the law of sines which is more easy. Now US = UW – SW = UC cosec - t = r cosec - t ………(e) And SUC = WSD = /2 - ……………………...(f) USQ = - (/2 - ) - = /2 + - ………………………....(g) From the law of sines,
- 7. β α 2 π sin UQ β sin US α 2 π sin SQ or, β α 2 π sin UQ β sin t α cosec r α 2 π sin x or, ) α β ( 2 π sin UQ β sin t α cosec r α cos x or, r sin cos sin sin cos( ) x t UQ cos cos sin sin UQ sin cos cos UQ r t by addendo or, UQ = x cos - t sin +r…………………………………..(h) and, QC= UC – UQ = r – UQ = t sin - x cos ………………(i)
- 8. But PQ lies in the plane of the circle and QC lies in the plane containing WUV ,and the two planes are perpendicular to each other. This point is important to understand, the plane of WSK is the plane of the paper and the plane of the intersection curve PSRT is perpendicular to the plane of the paper. , Hence PQ QC; so that PC2 = PQ2 + QC2, or, r2 = y2 + QC2 Or, r2 = y2 +( t sin - x cos )2……………………………………(j)
- 9. Now in the triangle WUC, r =PC = CU = UW sin = ( t + SU ) sin …..……(k) And in the triangle SUQ, α cos x α 2 π sin x β sin SU , i.e., SU = α cos β sin x So that r = ( t + SU ) sin =( t + α cos β sin x ) sin …..………….…(l) From (j) ( t + α cos β sin x )2 sin2 = y2 +( t sin - x cos )2…..…...(m) This is the relationship between x and y and as such, the equation to the curve of intersection we sought for.
- 10. Note that (j) could have been our required eqn. but it contains an extraneous variable r, which changed whenever x or y changed, so, naturally it had to be eliminated, and (m) is our equation sought for. Simplifying (m) we have, α sin α cos β sin x t y ) β cos x α sin t ( 2 2 2 2
- 11. or, α sin β sin x α cos t α cos y α cos ) β cos x α sin t ( 2 2 2 2 2 2 ………..(n) Though this is the equation of the ellipse, it needs to be further simplified and needs to be interpreted. On simplification we have, 2 2 2 2 2 (cos cos sin sin ) x 2 cos sin xt (cos cos sin sin ) +𝑦2 cos2 𝛼 = 0 or, 2 cos( )cos( ) x 2 2 2 cos sin cos( ) cos 0 xt y ………(o)
- 12. Considering the Δ SKT, we have, KST = , SKT = /2 + , STK = - KST - SKT = /2 – (+ ) and SK = 2SD = 2t sin , and call the maximum span of the curve of intersection , ST = 2a. Applying the law of sines to the Δ SKT , sin sin sin KT SK ST STK SKT
- 13. α 2 π sin a 2 ) β α ( 2 π sin α sin t 2 β sin KT or, α cos a 2 ) β α cos( α sin t 2 β sin KT ………………………………(p) From this cos(+) = (t/a) cos sin ; which further simplifies (o) as, 2 sin cos cos( ) t x a 2 cos sin cos( ) xt 2 2 cos 0 y or, 2 2 2 cos sin cos( )( 2 ) cos 0 t x xa y a , which, on completing square , 2 2 2 2 cos sin cos( )( ) cos cos sin cos( ) t x a y a a t ……(q) Since the right hand side is a positive constant, and both the terms in left side are positive assuming || and || both less than /2, |y| would be maximum only when |x – a| is minimum or 0. Calling this maximum value of y as b, we have b2 cos2 = at cos sin cos ( - ). …………………………(r)
- 14. This reduces (q) to, 2 2 2 2 2 2 cos ( ) cos b x a y a 2 2 2 cos a b Or, 1 b y a ) a x ( 2 2 2 2 which is equation to an ellipse with its left vertex at the origin, and a, b being its semi major and minor axes. If the origin is shifted from (0, 0) to (a, 0) and the new coordinates be X and Y, X = x – a, and Y = y, then the equation reduces to 1 b Y a X 2 2 2 2 ………………………...(s) which is standard equation to the ellipse. Thus it is shown that a plane section of a right circular cone not parallel to its generator and not parallel to its axis is actually an ellipse. The value of e From (p), 2 sin cos cos( ) t a a
- 15. And from (r), 2 ( ) . sin cos b at cos So 2 2 2 2 2 sin 1 cos b e a ………………(t) after simplification. Note that the value of e does neither depend on t, the distance of its vertex from the vertex of the cone. It depends upon b/a of course. Any ellipse parallel to this one, bounded by the cone shall have the same e value as it was expected. Taking a particular value of e, if we regard eqn.(t) as simply an equation in two variables and it is evident that a change in value of results in a corresponding change in the value of to maintain constancy of e. In other words, for the same ellipse, if the angle is changed, the cone is changed by change in ; that means, infinite number of cones can pass through the same ellipse, with different vertical angles. Is it just possible that orbits of all the ten planets of the Sun might be bounded by a common cone ?(enquire). It may be further observed that any plane section of a cone not parallel to its generator(s) and not parallel to its axis, shall result in an ellipse (cases of circles and degenerate point ellipses
- 16. at the vertex of cone included). Any plane section of a cone parallel to its generator shall result in a parabola (case of degenerate coincident st.lines passing through generators of cone included). Any plane section of a cone parallel to its axis shall result in a hyperbola (case of two st.lines passing through the vertex of cone included).These facts would be verified later on.