A variable point P is always equidistant from point (2,1) and the line y=3
Find the midpoint of intersecting points of the locus of P and the line 3x-y+4=0
the answer 指locus條equation係
(x-2)^2 x (y-1)^2 =(3-y)^2
但點解唔係(x-2)^2 x (y-1)^2 =1^2
所以我想追問寫呢種locus 既equation 有咩rules?
thank u very much!!!!!!!!!!!!!!!!!
Locus maths Thanku vy much?
回答 (1)
2016-02-04 4:23 pm
做呢種題形首先要由題目到搵番個關係先
然後再係個關係到寫番個 equation 出黎
"A variable point P is always equidistant from point (2,1) and the line y = 3"
關係：P 與 (2,1) 的距離 = P 與 y = 3 的垂直距離
設 P 為 (x,y)
√[ (x - 2)² + (y - 1)² ] = √[ (x - x)² + (y - 3)² ]
(x - 2)² + (y - 1)² = (3 - y)²
x² - 4x + 4 + y² - 2y + 1 = 9 - 6y + y²
y = - x²/4 + x + 1
Find the "midpoint" of intersecting points
{ y = - x²/4 + x + 1
{ 3x - y + 4 = 0
{ y = - x²/4 + x + 1 ...... ①
{ y = 3x + 4 ...... ②
① = ②
- x²/4 + x + 1 = 3x + 4
x² - 4x - 4 + 12x + 16 = 0
x² + 8x + 12 = 0
x-coordinate of the midpoint = - (8/1)/2 = -4
y-coordinate of the midpoint = 3(-4) + 4 = -8
∴ The midpoint = (-4,-8)
然後再係個關係到寫番個 equation 出黎
"A variable point P is always equidistant from point (2,1) and the line y = 3"
關係：P 與 (2,1) 的距離 = P 與 y = 3 的垂直距離
設 P 為 (x,y)
√[ (x - 2)² + (y - 1)² ] = √[ (x - x)² + (y - 3)² ]
(x - 2)² + (y - 1)² = (3 - y)²
x² - 4x + 4 + y² - 2y + 1 = 9 - 6y + y²
y = - x²/4 + x + 1
Find the "midpoint" of intersecting points
{ y = - x²/4 + x + 1
{ 3x - y + 4 = 0
{ y = - x²/4 + x + 1 ...... ①
{ y = 3x + 4 ...... ②
① = ②
- x²/4 + x + 1 = 3x + 4
x² - 4x - 4 + 12x + 16 = 0
x² + 8x + 12 = 0
x-coordinate of the midpoint = - (8/1)/2 = -4
y-coordinate of the midpoint = 3(-4) + 4 = -8
∴ The midpoint = (-4,-8)
收錄日期: 2021-04-18 14:24:33
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