"The straight line y=x+k intersects the curve y=x square at two pointsP and Q. Find the locus of the mid-point of PQ as k varies."
We should solve P and Q in terms of k first. But how can I do this?What should I do in order to deal with this type of locus straight linequestion?
F.5 A.Maths question (locus)
2009-04-11 5:58 am
回答 (2)
2009-04-11 6:06 am
✔ 最佳答案
y = x + k and y = x^2,
so x^2 - x - k = 0
sum of roots = 1/2.
Let the 2 intersecting points be a and b, so x co-ordinate of mid-point is (a + b)/2 = sum of roots/2 = (1/2)(1/2) = 1/4.
so locus of mid-point is x = 1/4.
2009-04-10 22:10:18 補充：
a is the x co-ordinate of P and b is the x co-ordinate of Q, so x co-ordinate of mid-point of P and Q will be (a + b)/2. But a and b are roots of equation x^2 - x - k = 0.
2009-04-11 12:56 pm
y = x + k
y = x^2
x^2 - x - k = 0
Sum of roots = -(-1)/1 = 1
The x co-ordinate of the mid-point of PQ = 1/2
The lowest point is (1/2, 1/4).
Thus, the locus of the mid-point is x = 1/2 and y >= 1/4.
y = x^2
x^2 - x - k = 0
Sum of roots = -(-1)/1 = 1
The x co-ordinate of the mid-point of PQ = 1/2
The lowest point is (1/2, 1/4).
Thus, the locus of the mid-point is x = 1/2 and y >= 1/4.
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