如图11,抛物线和直线y=kx-4k(k
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如图11,抛物线和直线y=kx-4k(k<0)与x轴,y轴相交于A,B两点,抛物线与x轴的另一个交点为C,抛物线的对称轴与x轴交于D点
(1)若点C的坐标为(-10,0)且tan∠DBO=3/4.求抛物线的解析式.
(2)若对称轴是直线x=-1,且∠ABD=90°.求抛物线的解析式.
(1)若点C的坐标为(-10,0)且tan∠DBO=3/4.求抛物线的解析式.
(2)若对称轴是直线x=-1,且∠ABD=90°.求抛物线的解析式.
没见有图,所以按叙述做.如不一致,请发图.
(1)
直线 y = kx - 4k = k(x - 4)
A(4,0),B(0,-4k)
C(-10,0),对称轴x = (-10 + 4)/2 = -3
y = a(x + 10)(x - 4)
x = 0,y = -40a = -4k,k = 10a
tan∠DBO = 3/4 = DO/OB = 3/(-4k)
k = -1
a = -1/10
抛物线的解析式:y = -(x + 10)(x - 4)/10 = -x²/10 - 3x/5 + 4
(2)
对称轴是直线x= -1
C(c,0)
-1 = (c + 4)/2
c = -6
y = a(x + 6)(x - 4)
x = 0,y = -24a = -4k,k = 6a
AB的斜率为k,BD的斜率为-1/k = (-4k - 0)/(0 + 1) = -4k
k² = 1/4
k = -1/2 (舍去k = 1/2 > 0)
a = k/6 = -1/12
y = -(x + 6)(x - 4)/12 = -x²/12 - x/6 + 2
(1)
直线 y = kx - 4k = k(x - 4)
A(4,0),B(0,-4k)
C(-10,0),对称轴x = (-10 + 4)/2 = -3
y = a(x + 10)(x - 4)
x = 0,y = -40a = -4k,k = 10a
tan∠DBO = 3/4 = DO/OB = 3/(-4k)
k = -1
a = -1/10
抛物线的解析式:y = -(x + 10)(x - 4)/10 = -x²/10 - 3x/5 + 4
(2)
对称轴是直线x= -1
C(c,0)
-1 = (c + 4)/2
c = -6
y = a(x + 6)(x - 4)
x = 0,y = -24a = -4k,k = 6a
AB的斜率为k,BD的斜率为-1/k = (-4k - 0)/(0 + 1) = -4k
k² = 1/4
k = -1/2 (舍去k = 1/2 > 0)
a = k/6 = -1/12
y = -(x + 6)(x - 4)/12 = -x²/12 - x/6 + 2
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