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第 17 週 · 坐標幾何 · 第 6 節

軌跡(locus)+ 本週關卡

軌跡就是「滿足某個條件的所有點」拼成的圖形。到定點等距是圓、到兩點等距是垂直平分線、到一條線等距是平行線 —— 把條件翻成方程就求出來了。本節講透軌跡,再以本週 BOSS 總關卡收官。配滿坐標圖例。

▶ 第一關: 軌跡是什麼 · 三種常見軌跡
locus = all points that satisfy a conditionAeach dot is dist 5 from Anot on locus(wrong dist)
A(2,3)r = 5locus: circle, centre A, radius 5
ABperpendicular bisectorsame dist to A and B
x-axisy = 4 (dist 4 above)y = −4 (dist 4 below)
▶ 第二關: 求軌跡方程(設 → 翻譯 → 化簡)
1. set pointP(x, y)2. translatecondition → eqn3. simplifytidy to simplestfinding a locus equation
set P(x,y), dist to A(2,3) is 5sqrt((x-2)^2 + (y-3)^2) = 5square both sides(x - 2)^2 + (y - 3)^2 = 25circle, r = 5
P(x,y) equidistant from A(0,0), B(4,0)x^2 + y^2 = (x - 4)^2 + y^2x^2 + y^2 = x^2 - 8x + 16 + y^20 = -8x + 16x = 2 (perpendicular bisector)
▶ 第三關: 軌跡方程板書鞏固
(x - 2)^2 + (y - 3)^2 = 25centre (2, 3)radius = sqrt(25) = 5centre sign flips; radius is sqrt of RHS
A(1,0)B(5,0)mid(3,0)x = 3
⚔ BOSS: W17 坐標幾何 · 總關卡
BOSS: five W17 skills across three questionsQ13 line eqn + parallely = 2x + c, thru (0,5) -> c = 5Q14 circle eqn + pointd = sqrt(9+16) = 5 = r -> onQ15 line and circle (tangent)d = r = 5 -> tangent (touch once)line eqn | parallel/perp | circle eqn | point&circle | line&circleclear all three to finish week 17