係 Maths Core 我地成日都會講 parabola,但究竟 parabola 係咩黎既呢?我地經常話 y=ax2+bx+c (where a, b and c are real constants with a≠0 既 graph 係一條 parabola,但…… 其實我地開口埋口講既 parabola 又係咩黎? 其實數學裡面有一種 curve 叫做 conic sections,即係係一個 double infinite right circular cone 用一塊 plane 黎 cut 出黎既 curve,就好似下圖咁:
Figure 1: A figure showing different types of conic sections
[Source: Calculus by Douglas F. Riddle, Wadsworth, Inc. 1984.]
當中最後一個 case 就係個 plane 岩岩好 touch 到個 cone 係一條 straight line(下右一)上面,而將呢個 plane translate outward 少少就會出 parabola(上右一)喇!當然,我地 formal d 講既話,就會係個 plane 同 vertical plane 形成既角同個 cone 既 semi-vertex angle 係 equal(semi-vertex angle 即係個 cone 係 apex 道形成既最大既角 ÷2)
但呢個 parabola 又有d 咩特別?點解其他既 graph,比如 y=x4 既 graph 點解就唔係 parabola?
Figure 2: A figure showing the graph of y=x4
咁我地就要 investigate 下 parabola 既 properties 喇!
Figure 3: A Figure showing the idea of the situation
係上面個圖道,O 係 double cone 既 apex,Π 就係一個同 vertical plane 形成既角同 semi-vertex angle 一樣既 plane,所以就可以 cut 成一條 parabola p,裡面既 sphere 就 armarm 好畫到同個 cone(Π 向住 O 既 half plane 個邊)touch 而且亦同 Π touch 係 point F 既 plane,個 sphere 同個 double cone 既 intersection 係一個 circle c,而個 circle lie on 既 plane 我地叫佢做 Ω
如果 P 係 p 上面既一個 moving point,而 A 就係 OP 同 c 既 intersection,B 係 P 係 Ω 上面既 projection,C 係 P 係 Π 同埋 Ω 既 intersecting line(denoted by l)上面既 projection
由我地係 3D solids 有關 projection 既知識話比我地聽,∠PBA=∠PBC=90∘。
另外,∠PAB 同 ∠PCB 都係 semi-vertex angle (by the definition of parabola),所以:
∠PAB=∠PCB(proved)∠PBA=∠PBC(proved)PB=PB(common side)△PAB≅△PCB(A.A.S.)PA=PC(corr. sides, ~△s)
另外,∠PAB 同 ∠PCB 都係 semi-vertex angle (by the definition of parabola),所以:
∠PAB=∠PCB(proved)∠PBA=∠PBC(proved)PB=PB(common side)△PAB≅△PCB(A.A.S.)PA=PC(corr. sides, ~△s)
由我地既 tangent properties 可以知道 PF=PA,因此我地就知道 PF=PC,即即我地搵到 Π (parabola p 屬於既 plane)上面既一點 F 同埋直線 l 令到 parabola p 上面既任何一點同 F 以及同 l equidistant,因此呢個就係 parabola 另一個 definition(當然實際上都要調轉再 prove 一次,但由於時間關係,你地都可以自己試下架):
A parabola is a curve such that there exists a point F and a line l such that every point P lying on the parabola is equidistant to F and to l.
A parabola is a curve such that there exists a point F and a line l such that every point P lying on the parabola is equidistant to F and to l.
咁呢句又點樣帶番落我地平時既 quadratic graph 道啊?咁我地就要用 locus 既方法喇!頭先上面個句正正就係 locus 既句子黎。所以我地就可以 let P=(x,y) 先,而我地就想個 l 係 horizontal,所以我地就 set P=(a,b) 同埋 y=k,where k≠b, a, b and k are real constants。
√(x−a)2+(y−b)2=√(x−x)2+(y−k)2(x−a)2+(y−b)2=(y−k)2(2b−2k)y=(x−a)2+b2−k2y=12b−2k(x2−2ax+a2+b2−k2)(∵
咁我地就出到條 parabola 喇!!!
但…… 點解我地叫 parabola 做拋物線既?咁我地就要用下 F.4 上學期既 mechanics 喇!
係 projectile motion (當然係 without air resistance 啦) 裡面,如果一個 object 由 (0,0) 呢個 point 開始擲出,initial velocity 係 u,inclination 係 \theta,咁我地就有 \begin{cases} x = ut \cos\theta \\ y = ut \sin\theta - \frac{1}{2} gt^{2}\end{cases}.
√(x−a)2+(y−b)2=√(x−x)2+(y−k)2(x−a)2+(y−b)2=(y−k)2(2b−2k)y=(x−a)2+b2−k2y=12b−2k(x2−2ax+a2+b2−k2)(∵
咁我地就出到條 parabola 喇!!!
但…… 點解我地叫 parabola 做拋物線既?咁我地就要用下 F.4 上學期既 mechanics 喇!
係 projectile motion (當然係 without air resistance 啦) 裡面,如果一個 object 由 (0,0) 呢個 point 開始擲出,initial velocity 係 u,inclination 係 \theta,咁我地就有 \begin{cases} x = ut \cos\theta \\ y = ut \sin\theta - \frac{1}{2} gt^{2}\end{cases}.
由第一條 equation 可知,t=\frac{x}{u\cos\theta},sub. 番落第二條式就見到 y=x\tan\theta - \frac{gx^{2}}{2\cos ^{2}\theta},而因為 g、u 同埋 \theta 係拋左個 object 出去已經係 constant,所以 y against x 既 graph 就會係 parabola,所以 parabola 既中文名就係「拋物線」喇!
今日講到咁多,希望大家都會知道大家學緊既 parabola 其實係點黎啦~