maths locus

2014-10-26 6:34 am
The figure shows the straight line L:y= -1/2 and a point A(1/2,0).
It is given that a moving point P is equidistant from A and L.
a) Find the equation of the locus of P.
b) If D(-3,d) lies on the locus in (a), find the value of d.


The figure: http://postimg.org/image/5ae9o27gj/
更新1:

修正: The figure: http://postimg.org/image/sd1y3ynrd/

回答 (1)

2014-10-26 7:13 am
✔ 最佳答案
a)Let (x,y) be P:
The distance of P from A = √[(x - 1/2)² + (y - 0)²]= √[(x - 1/2)² + y²].
The distance of P from the line L = the distance of P from the point (x , - 1/2)
√[(x - x)² + (y + 1/2)²] = √(y + 1/2)².By given ,
√( (x - 1/2)² + y² ) = √(y + 1/2)²
(x - 1/2)² + y² = (y + 1/2)²
x² - x + 1/4 + y² = y² + y + 1/4
x² - x - y = 0The equation of the locus of P is x² - x - y = 0.
b)Sub. x = - 3 and y = d to the equation of the locus of P :
(- 3)² - (-3) - d = 0
d = 12

2014-10-25 23:19:06 補充:
注意 √(y + 1/2)² 不可寫作 y + 1/2 ,
例如 √(- 5)² = - 5 是錯的。

2014-10-25 23:28:27 補充:
如你學過絕對值, √(y + 1/2)² 可寫作 | y + 1/2 | ,
於是可直接寫 The distance of P from the line L = | y + 1/2 | 。

2014-10-25 23:32:57 補充:
之後運算就是:
√( (x - 1/2)² + y² ) = | y + 1/2 |²
(x - 1/2)² + y² = (y + 1/2)²
............


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