Lecture 1.16 is parabola really a section of a cone?
- 1. To show that a parabola is actually a section of a cone: Let a St. Line WS revolve around a fixed st. line WG always making an angle with it at W and generate a right circular cone Evidently a circle is drawn by WS where SK is its diameter with its center at D as shown , the circle formed by intersection of a plane perpendicular to WG, the axis of the cone. From congruence of triangles WSD and WDK having a common side and two angles (half angle of the cone at its vertex) and /2 equal, it follows that SD = DK = WS sin or t sin , if WS = t. Let another plane cuts the cone at S which is parallel to the line WK and is perpendicular to the plane of the paper or perpendicular to the plane WSK. It may be observed that TST’ is the boundary of the surface generated by cutting of the cone by the plane concerned. Let P be any point on the curve of intersection TST’ with cone which we are considering. Take a st. line SX in our intersecting plane perpendicular to the plane of WSK, the line SX being parallel to WH . Drop a perpendicular PQ from P onto SX and extend it to meet the curve PST at R. Set up a rectangular Cartesian coordinate system with SX as x-axis, origin at S and SY as y-axis perpendicular to SX, in the plane of PSRT. The coordinates of the point P are x = SQ and y = PQ.
- 2. Take a plane containing the line PQ and perpendicular to WG and perpendicular to the plane of the paper. Its intersection with the cone is a circle UPVR with center C say, on WG and a diameter UV on the plane WSK, bounded by the cone; i.e., the points S and K lie on the cone. Now QC, a part of UV in the plane of the circle PUV and PQ is in the plane of the curve PST and the two planes are perpendicular to each other ,it follows that PQ QC so that PQ2 + QC2 = PC2 or, y2 + QC2 = r2 ………………..(a)
- 4. Where PQ = y, and PC is radius of the circle PURV , say, r. If we could relate QC with x or SQ, this would serve as a relationship between x and y, hence it would be equation to the curve of intersection we are seeking for. Of course r has to be eliminated as it is a variable changing as the point P moves along the curve PST. Now QC is UC – UQ ; so that we have to concentrate on the triangle SUQ. We immediately note that SUQ = WSD = /2 - and SQU = KVQ = WKD = /2 - ; (as SX || WH and UV || SK, the triangles WSK and SUQ are similar). Similarly if SL is drawn perpendicular to UQ, then USL = SWC = ; and USL = LSQ = , from congruence of the triangles SUL and SLQ. Also SU = SQ = x and UL = LQ = SQ sin = x sin ………………….(b) From (a),y2 + QC2 = r2 y2 + (UC - 2UL)2 = UC2 y2 + (r – 2x sin )2 = r2 y2 + (UW sin – 2x sin )2 = UW2 sin2 y2 + {(x +t) sin – 2x sin }2 = (x +t)2 sin2 ,as,(UW=US+SW=SQ+SW=s+t) y2 + (t - x)2 sin2 = (x +t)2 sin2 y2 = (x + t)2 sin2 - (t - x)2 sin2 = 4xt sin2 This is the standard eqn. of a parabola y2 = 4ax…………………………………………………….(c) where t sin2 may be taken as ‘a’, the distance of the focus of the parabola
- 5. from its vertex S and evidently with value of e =1. This a should not be confused with semi-major axis of the ellipse. ( Putting T in place of t sin and X in place of x sin and Y in place of y, the eqn. becomes, (X - T)2 + Y2 = (X + T)2 ; thus the distance of the point (X, Y) from the fixed point ( T, 0) is equal to the distance of the point (X, Y) from the fixed st.line X = - T )