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Similar to SUEC 高中 Adv Maths (Locus) (Part 1).pptx
SUEC 高中 Adv Maths (Locus) (Part 1).pptx
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- 5. 𝒑𝒈 𝟗𝟎 𝑥1 ,𝑦1 𝑥2 , 𝑦2
- 6.
- 7. 𝒑𝒈 𝟗𝟏 𝜆 = 2 5 𝑐= 0 𝑎2 + 𝑏2 = 𝑟2 𝑥2 − 8𝑦 + 16 = 0 𝑥2 + 𝑦2 = 5 𝑥 = 𝑐
- 8. 𝑥2 − 8𝑦 +16 = 0 𝑙𝑒𝑡 𝑃 = (𝑥, 𝑦) 𝑥 𝑎𝑥𝑖𝑠 = (𝑥, 0) 𝑃𝐹 = 𝑃 𝑥 𝑎𝑥𝑖𝑠 𝐹 (0, 4) 𝑥2 + 𝑦 − 4 2 = 𝑦2 𝑥2 + 𝑦2 − 8𝑦 + 16 = 𝑦2 𝑥2 + 𝑦2 = 5 𝑃𝐴 2 + 𝑃𝐵 2 = 42 𝑙𝑒𝑡 𝑃 = (𝑥, 𝑦) 𝑥 − 4 2 + 𝑦2 + 𝑥 + 4 2 + 𝑦2 = 42 𝑥2 + 16 − 8𝑥 + 𝑦2 + 𝑥2 + 16 + 8𝑥 + 𝑦2 = 42
- 9. 2𝑐𝑥 = 2𝑐2 𝑂𝑃2 − 𝐴𝑃 2 = 𝑐2 𝑙𝑒𝑡 𝑃 = (𝑥, 𝑦) 𝑥2 + 𝑦2 − 𝑥 − 𝑐 2 + 𝑦2 = 𝑐2 𝑥2 + 𝑦2 − 𝑥2 + 𝑐2 − 2𝑐𝑥 + 𝑦2 = 𝑐2 𝑥 = 𝑐 3𝑥2 + 3𝑦2 − 2𝑥 − 20𝑦 + 27 = 0 𝑙𝑒𝑡 𝑃 = (𝑥, 𝑦) 𝐴 = (3,2) 𝐵 = (1,3) 𝑃𝐴 = 2𝑃𝐵 𝑥 − 3 2 + 𝑦 − 2 2 = 2 𝑥 − 1 2 + 𝑦 − 3 2 𝑥2 + 9 − 6𝑥 + 𝑦2 + 4 − 4𝑦 = 4 (𝑥2 + 1 − 2𝑥 + 𝑦2 + 9 − 6𝑦 𝑥 − 3 2 + 𝑦 − 2 2 = 4 𝑥 − 1 2 + 𝑦 − 3 2
- 10. 𝒑𝒈 𝟗𝟏 3𝑥2 + 3𝑦2 −2𝑥 − 20𝑦 + 27 = 0 𝑥2 + 𝑦2 = 4 𝑥2 + 𝑦2 − 8𝑥 − 8𝑦 + 7 = 0
- 11. 𝑙𝑒𝑡 𝑃 =(𝑥, 𝑦) 𝑚𝐴𝑃 × 𝑚𝐵𝑃 = −1 𝑦 𝑥 − 1 × 𝑦 − 8 𝑥 − 7 = −1 𝑦2 − 8𝑦 = − 𝑥2 + 7 − 8𝑥 𝒑𝒈 𝟗𝟏 𝑥2 + 𝑦2 − 8𝑥 − 8𝑦 + 7 = 0
- 12. 𝒑𝒈 𝟗𝟐 𝑐𝑒𝑛𝑡𝑒𝑟 &𝑟𝑎𝑑𝑖𝑢𝑠
- 13. 𝒑𝒈 𝟗𝟒 𝑥 −3 2 + 𝑦 + 2 2 = 36 𝑥2 + 𝑦2 = 25 9 𝑥 − 8 2 + 𝑦 + 3 2 = 25 𝑥 − 𝑎 2 + 𝑦 + 𝑏 2 = 𝑎 + 𝑏 2
- 14. 𝒑𝒈 𝟗𝟐 −𝟗𝟑 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝐴𝐵 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
- 15. 𝒑𝒈 𝟗𝟑 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ𝐴𝐵 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
- 16.
- 17.
- 18.
- 19. 𝒑𝒈 𝟗𝟒 𝑥 +6 2 + 𝑦 − 3 2 = 10 𝑥 − 5 2 + 𝑦 + 3 2 = 13 | 𝑃𝑅 | = 5 2
- 20. 𝒑𝒈 𝟗𝟒 | 𝑃𝑅| = 5 2 𝑅 𝐶 = 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑃𝑄 = (5, −3) 𝑟 = 13
- 21. 𝒑𝒈 𝟗𝟒 𝑥2 + 𝑦− 5 2 = 34 𝑥 − 4 2 + 𝑦2 = 25 𝑥 + 5 2 + 𝑦 − 4 2 = 16 𝑥 − 1 2 + 𝑦 + 3 2 = 13 𝑥 − 18 2 + 𝑦 − 16 2 = 100 𝑥 − 2 2 + 𝑦 − 4 2 = 100
- 22. 𝒑𝒈 𝟗𝟒 𝑥 −18 2 + 𝑦 − 16 2 = 100 𝑟 = 𝑑 = 4𝑥 + 3𝑦 − 70 5 = 10 (10, 10) 𝑙1 ∶ 4𝑥 + 3𝑦 − 70 = 0 , 𝑚 = 4 3 𝑙2 ∶ 𝑦 − 10 = 3 4 (𝑥 − 10) 4𝑦 − 3𝑥 − 10 = 0 − −① 4𝑥 + 3𝑦 − 120 = 0 − −② 𝑜𝑟 4𝑥 + 3𝑦 − 50 = 0 𝑐𝑎𝑛 𝑏𝑒 𝑡ℎ𝑖𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑎𝑠 𝑤𝑒𝑙𝑙 𝑑𝑢𝑒 𝑡𝑜 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑎𝑙𝑢𝑒 𝑠𝑜𝑙𝑣𝑒 𝑠𝑦𝑡𝑒𝑚𝑠 𝑜𝑓 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑓𝑜𝑟 ① 𝑎𝑛𝑑 ②
- 23. 𝒑𝒈 𝟗𝟒 𝐴 𝐵 𝐶 𝐴𝐵 =𝐴𝐶 = 𝑟 = 10 ∴ 𝐴 = 𝑐𝑒𝑛𝑡𝑒𝑟 = 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐵𝐷 𝐷 = (5, −1)
- 24.
- 25.
- 26.
- 27.
- 28. 𝒑𝒈 𝟗𝟖 3𝑥2 +3𝑦2 + 24𝑥 + 82𝑦 = 0 𝑥2 + 𝑦2 − 𝑏𝑥 − 𝑎𝑦 = 0 25𝑥2 + 25𝑦2 − 277𝑥 − 83𝑦 − 278 = 0 𝑥2 + 𝑦2 − 8𝑥 − 2𝑦 + 12 = 0
- 29. 𝒑𝒈 𝟗𝟔 2 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑝𝑜𝑖𝑛𝑡𝑠
- 30. 𝒑𝒈 𝟗𝟔 2 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑝𝑜𝑖𝑛𝑡𝑠 𝐶 𝐵 𝐴
- 31. 𝒑𝒈 𝟗𝟕 −𝟗𝟖 𝐼𝑡 𝑖𝑠 𝑘𝑛𝑜𝑤𝑛 𝑡ℎ𝑎𝑡 𝑎 𝑐𝑢𝑟𝑣𝑒 𝑖𝑠 𝑡ℎ𝑒 𝑙𝑜𝑐𝑢𝑠 𝑜𝑓 𝑎 𝑝𝑜𝑖𝑛𝑡 0 (0, 0) 𝑎𝑛𝑑 𝐴 (3, 0) 𝑤ℎ𝑜𝑠𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑟𝑎𝑡𝑖𝑜 𝑖𝑠 1 2 . 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑖𝑠 𝑐𝑢𝑟𝑣𝑒 𝑎𝑛𝑑 𝑠ℎ𝑜𝑤 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ.
- 32. 𝒑𝒈 𝟗𝟖 2𝑥2 + 2𝑦2 −11𝑥 − 19𝑦 + 44 = 0 𝑥2 + 𝑦2 − 𝑎𝑥 − 𝑏𝑦 = 0 𝑥2 + 𝑦2 − 8𝑥 − 9 = 0 𝑥2 + 𝑦2 − 2𝑥 = 0 2𝑥2 + 2𝑦2 − 21𝑥 + 8𝑦 + 60 = 0
- 33. 𝒑𝒈 𝟗𝟖 2𝑥2 + 2𝑦2 −11𝑥 − 19𝑦 + 44 = 0 𝐶 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡 𝑥 + 𝑦 = 4 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑦 − 𝑎𝑥𝑖𝑠 𝐶 = (0, 4) 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝐴, 𝐵, 𝐶 𝑖𝑛𝑡𝑜 𝐴 ∶ 4 + 4 + 4𝑔 + 4𝑓 + 𝑐 = 0 8 + 4𝑔 + 4𝑓 + 𝑐 = 0 − −① 𝐵 ∶ 25 + 9 + 10𝑔 + 6𝑓 + 𝑐 = 0 34 + 10𝑔 + 6𝑓 + 𝑐 = 0 − −② 𝐶 ∶ 16 + 8𝑓 + 𝑐 = 0 − −③
- 34. 𝒑𝒈 𝟗𝟖 ∴ 𝑥2 +𝑦2 −𝑎𝑥 − 𝑏𝑦 = 0 𝐿𝑒𝑡 𝐴 & 𝐵 = 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑥 & 𝑦 𝑎𝑥𝑒𝑠. 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑎𝑙𝑙 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑖𝑛𝑡𝑜 𝑜𝑟𝑖𝑔𝑖𝑛 ∶ 𝑐 = 0 𝐴 ∶ 𝑎2 + 2𝑎𝑔 = 0 𝐵 ∶ 𝑏2 + 2𝑏𝑓 = 0 𝐴 = (𝑎, 0) , 𝐵 = (0. 𝑏) 2𝑔 = − 2𝑓 = −𝑏
- 35.
- 36. 𝒑𝒈 𝟗𝟖 𝑥2 +𝑦2 = 𝑎2 𝐿𝑒𝑡 𝑃 = (𝑥, 𝑦) = 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐴𝐵 ∴ 𝐴 = (2𝑥, 0), 𝐵 = (0, 2𝑦) 𝐴𝐵 = 2𝑥 2 + 2𝑦 2 = 2𝑎 2𝑥 2 + 2𝑦 2 = 4𝑎2
Editor's Notes
- #2 https://www.onlinemathlearning.com/locus-math.html
- #9 5. Knowing that the distance from point P to the x-axis is equal to its distance from point F (0, 4), find the orbital equation of point P. 6. Given that the sum of the squares of the distance between point P and point A (4,0) and point B (-4, 0) is 42. find the orbital equation of point P.
- #10 9. The distant from a moving point to point (3,2) is equal to twice the distance to point (1,3).
- #12 The central angle APB formed by a moving point P and 2 fix points A (1,0) & B (7,8) is always a right angle.
- #23 r = 10, tangent to line 4x+3y-70 = 0 at point (10,10). Find eq
- #24 The coordinates of vertex A of an isosceles triangle are (4,2) and the coordinates of one endpoint B of the base (3.5). Find the eq at the other endpoint C and explain.
- #34 A circle passes through the points A (2, 2), B (5, 3), and passes through the intersection of the line x + y = 4 and the y-axis, find the equation of this circle.
- #35 A circle passed through the origin of coordinates and the intercepts on the x-axis and y-axis are a and b respectively. Find the equation of this circle.
- #37 A line segment | AB | = 2a, its two endpoints A and B slide on the x-axis and y-axis respectively, find the orbit of the midpoint P of AB.