hoikit asked in 科學及數學數學 · 1 decade ago

求 New Progress in certifficate mathatics revised edition 20點

求 New Progress in certifficate mathatics revised edition as Re.ex 9

Q18,19, answer 超急,快,要有solution

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  • 1 decade ago
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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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