標題:
求 New Progress in certifficate mathatics revised edition 20點
發問:
求 New Progress in certifficate mathatics revised edition as Re.ex 9 Q18,19, answer 超急,快,要有solution
最佳解答:
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let the slope of L = m , then the equation of L is : (y-3)=m(x+3) y=mx+3(m+1) when y=0, x= -3(m+1)/m let the angle = x then m= tan(90+x)=-cot(x) slope of R= tan(90-x)=cot (x)=-m so.. the equation of R = (y-0)=-m[x+3(m+1)/m] y=-mx-3(m+1) put it into C, x^2+[-mx-3(m+1)]^2-4x-4[-mx-3(m+1)]+7=0 Dlata=0 then we get m=-3/4 or -4/3.. so the equation is 3x+4y-3=0 and 4x+3y+3=0 2008-05-26 11:39:44 補充： 19. let the center is (x, 1-2x), abs[(x+2-4x+5)/√5]=√[(x-1)^2+(2-2x)^2] (-3x+7)^2=5(5x^2-10x+5) 9x^2-42x+49=25x^2-50x+25 16x^2-8x-24=0 2x^2-x-3=0 (2x-3)(x+1)=0 x=3/2 or x=-1 y=-2 , y=3, radius = √[(-1-1)^2+(2+2)^2]=√20 so the equation are : C1:(x+1)^2+(y-3)^2=20 and C2: (x-3/2)^2+(y+2)^2=20
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