如图,抛物线y=ax²+bx=c(a>0)与y轴交与点C,直线L,经过点C且平行与x轴,将L1向上平移t个单位

2个回答

  • 1.C(0,1)

    L2:y = 3

    y = x²/2 -3x/2 + 1 =3

    x² - 3x -4 = 0

    (x-4)(x+1) = 0

    A(-1,3),B(4,3)

    AC的斜率k1 = (3-1)/(-1-0) = -2

    BC的斜率k2 = (3 - 1)/(4 - 0) = 1/2

    k1 * k2 = -1

    ΔABC为直角三角形

    2.l1:y = c

    l2:y = c + t (t > 0)

    l2与抛物线的交点为A、B,ax²+bx + c = c + t

    ax²+bx - t =0

    A,B的坐标为

    A([-b - √(b²+4at)]/(2a),c + t)

    B([-b + √(b²+4at)]/(2a),c + t)

    C(0,c)

    AC的斜率k1 = (c+t-c)(2a)/[-b - √(b²+4at)] = -2at/[b + √(b²+4at)]

    BC的斜率k2 = (c+t-c)(2a)/[-b + √(b²+4at)] = 2at/[-b + √(b²+4at)]

    k1*k2 = -1

    4a²t² = 4at

    at = 1

    t = 1/a

    3.抛物线的对称轴:x = -b/(2a)

    A关于y轴的对称点A' 的坐标为([b + √(b²+4at)]/(2a),c + t)

    |A'B| = [-b + √(b²+4at)]/(2a) - [b + √(b²+4at)]/(2a) = -b/a

    a > 0,显然 b < 0

    A' 在抛物线的对称轴上,-b/(2a) = [b + √(b²+4at)]/(2a)

    -2b = √(b²+4at)

    3b² = 4at = 4

    b = -√(4/3) (正值舍去)

    y = ax²+bx + c =c

    ax²+bx = 0

    x(ax +b) = 0

    x = 0(点C),x = -a/b

    D(-a/b,c)

    |CD| = -a/b

    AB,CD平行

    |CD| = |A'B|,四边形A'CDB为平行四边形,CD上的高h = c+t -c = t

    四边形A'CDB的面积S = |CD|*h = -(a/b)t = -at/b = -1/b = 1/√(4/3) = (√3)/2