Cho (P): y=x^2/2 va duong thang (d): y=mx+1/2 a) Ve (P) b) CM: voi moi m duong thang (d) luon di qua mot diem co dinh c) CM: voi moi m (d) luon cat (P) tai hai diem phan biet
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19 tháng 11 2017
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11 tháng 3 2018
a)Xét (O) có, ^AMB=^ANB=^NBM=^NAM=90 độ ( góc nội tiếp chắn nửa đt)
Xét tứ giác ANBM có : ^AMB=^ANB=^NBM=90 độ (cmt)
=> TG ANBM là hcn
HD
17 tháng 12 2017
1) gọi đường thẳng cần tìm là y=ax+b(d1)
vì đt d1 vuông góc vs đt y=2x-1 nên:
a.2=-1 <=> a= \(\dfrac{-1}{2}\)
vì đt d1 đi qua điểm M (-1;1) nên ta có pt:
1=\(\dfrac{-1}{2}\) .(-1)+b <=> b=\(\dfrac{1}{2}\)
Vậy h/s cần tìm là y=\(\dfrac{-1}{2}\) x+\(\dfrac{1}{2}\)
2) gọi đường thẳng cần tìm là y=ax+b(d)
vì đt d // đt y=3x+1 nên:
a=3
vì đt d cắt trục tung tại điểm có tung độ bằng 4 nên : b=4
vậy h/s cần tìm là y=3x+4
3) đk :m\(\ne\)2
vì đt y=2x-1 cắt tại tung độ tại điểm có tung độ bằng -x nên ta có pt :
-x=2x-1 <=> x=\(\dfrac{1}{3}\)
Ta có đt y=mx+1 cắt tại tung độ tại điểm có tung độ bằng -x nên ta có pt :
-\(\dfrac{1}{3}\) =m.\(\dfrac{1}{3}\) +1 <=> m=-4 (tmđk )
Vậy để y=mx+1 va y=2x-1 cắt nhau tại điểm thuộc y=-x thì m= -4
