clevertt
幼苗
共回答了18个问题采纳率:88.9% 举报
(3)双曲线 y = -4/x
抛物线y = -x² - 3x = -(x + 3/2)² + 9/4
E(-3/2,9/4)
△ABD和△ABE的底AB相同,只需D与AB的距离H为E与AB的距离h的8倍
A(-2,2),B(1,-4)
AB的解析式:(y + 4)/(2+ 4) = (x - 1)/(-2 - 1)
2x + y + 2= 0
h = |2(-3/2) + 9/4 + 2|/√5 = √5/4
H = 8h = 2√5
设D(d,-d² - 3d)
H = |2d - d² - 3d + 2|/√5 = |d² + d -2|/√5 = 2√5
|d² + d -2| = 10
d² + d -2 = -10 (无解)
或d² + d-2 = 10
d² + d -12 = 0
(d+4)(d-3) = 0
d = -4或d = 3
D(-4,-4)或D(3,-18)
1年前
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h = |2(-3/2) + 9/4 + 2|/√5 = √5/4和H = |2d - d² - 3d + 2|/√5 = |d² + d -2|/√5 = 2√5 根号五是怎么来的?
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clevertt
点A(a, b)与直线Ax + By + C =0的距离公式: d =|Aa+ Bb + C|/√(A² + B²) 这里A = 2, B = 1, √(A² + B²) = √5