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A.maths 高手請答
發問:
1.a line L: y=k(x-1) ,k is a real constant . L cuts the curve y^2 =4x at two points A(x1,y1) and B(x2,y2). x1 and x2 are the roots of k^2x^2 -2(k^2+2)x +k^2 =0 M is the mid -point of AB As k varies ,find the equation od the locus of M. 請列明steps ~~~ thz! 如果可以ge 話, plz 畫埋圖
最佳解答:
1.a line L: y=k(x-1) ,k is a real constant . L cuts the curve y^2 =4x at two points A(x1,y1) and B(x2,y2). x1 and x2 are the roots of k^2x^2 -2(k^2+2)x +k^2 =0 M is the mid -point of AB As k varies ,find the equation od the locus of M. ANSWER let the point M (a,b) Then a=(x1+x2)/2 ; b=(y1+y2)/2 But (x1+x2)=2(k^2+2)/k^2 [sum of the roots] So 2a=2(k^2+2)/k^2 or a=(k^2+2)/k^2 a=1+2/k^2 2/k^2=(a-1) k=√[2/(a-1)] Also y1=k(x1-1) ; y2=k(x2-1) y1+y2 =k(x1-1+x2-1) =k(x1+x2-2) (y1+y2)=k(x1+x2-2) 2b=k(2a-2) 2b=2k(a-1) b=k(a-1) substitute the value of k b={√[2/(a-1)]}(a-1) b^2=2(a-1)^2/(a-1) b^2=2(a-1) change back a,b to x,y The equation of the locus of M is y^2=2(x-1)
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