高中数学圆锥曲线问题求解;
1.已知双曲线X²-Y²/2=1,过点A(2,1)的直线与已知双曲线交于P,Q两点,求PQ重点的轨迹方程。2.已知椭圆X²/9+Y²...
1.已知双曲线X²-Y²/2=1,过点A(2,1)的直线与已知双曲线交于P,Q两点,求PQ重点的轨迹方程。
2.已知椭圆X²/9+Y²/4=1及过点D(2,1),过点D任意引直线交椭圆于A,B两点,求线段AB中点M的轨迹方程.
3.已知椭圆X²/16+Y²/9=1,F1,、F2分别为他的焦点,CD为过F1的弦,则△F2CD的周长为多少?
已知方程X²/(3+k)+Y²/(2-k)=1表示椭圆,则k的取值范围是多少?
已知椭圆Y²/9+X²=1,一条不与坐标轴平行的直线l与椭圆交于不同的点M,N,且线段MN的中点的横坐标为-1/2,求直线l的取值范围。 展开
2.已知椭圆X²/9+Y²/4=1及过点D(2,1),过点D任意引直线交椭圆于A,B两点,求线段AB中点M的轨迹方程.
3.已知椭圆X²/16+Y²/9=1,F1,、F2分别为他的焦点,CD为过F1的弦,则△F2CD的周长为多少?
已知方程X²/(3+k)+Y²/(2-k)=1表示椭圆,则k的取值范围是多少?
已知椭圆Y²/9+X²=1,一条不与坐标轴平行的直线l与椭圆交于不同的点M,N,且线段MN的中点的横坐标为-1/2,求直线l的取值范围。 展开
4个回答
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1.
设P(m, n), Q(p, q):
m² - n²/2 = 1
p² - q²/2 = 1
相减, (n-q)/(m - p) = 2(m+p)/(n+q) (i)
PQ的中点M((m+p)/2, (n+q)/2)
令M(x, y):
m + p = 2x (ii)
n + q = 2y (iii)
PQ的斜率k = (n - q)/(m - p)
MA的斜率k' = (y - 1)/(x - 2)
k = k', (x - 1)/(y - 2) = (n - q)/(m - p) = 2(m+p)/(n+q) (i)
= 4x/(2y) (ii)(iii)
= 2x/y
2x² - 4x - y² + y = 0
2.
与1类似
设P(m, n), Q(p, q):m²/9 + n²/4 = 1
p²/9 + q²/4 = 1
相减, (n-q)/(m - p) = -4(m+p)/[9(n+q)] (i)
PQ的中点M((m+p)/2, (n+q)/2)
令M(x, y):
m + p = 2x (ii)
n + q = 2y (iii)
PQ的斜率k = (n - q)/(m - p) = -4(m+p)/[9(n+q)]
MA的斜率k' = (y - 1)/(x - 2)
k = k'
(y - 1)/(x - 2) = -4*2x/(9*2y)
4x²- 8x + 9y² - 9y = 0
(3)
a = 4, 按椭圆定义,|F1C| + |F2C| = 2a = 8, |F1D| + |F2D| = 2a = 8
△F2CD的周长 = |F1C| + |F2C| + |F1D| + |F2D| = 16
3+k > 0, k > -3
2 - k > 0, k < 2
-3 < k < 2
最后一题不太清楚。
设P(m, n), Q(p, q):
m² - n²/2 = 1
p² - q²/2 = 1
相减, (n-q)/(m - p) = 2(m+p)/(n+q) (i)
PQ的中点M((m+p)/2, (n+q)/2)
令M(x, y):
m + p = 2x (ii)
n + q = 2y (iii)
PQ的斜率k = (n - q)/(m - p)
MA的斜率k' = (y - 1)/(x - 2)
k = k', (x - 1)/(y - 2) = (n - q)/(m - p) = 2(m+p)/(n+q) (i)
= 4x/(2y) (ii)(iii)
= 2x/y
2x² - 4x - y² + y = 0
2.
与1类似
设P(m, n), Q(p, q):m²/9 + n²/4 = 1
p²/9 + q²/4 = 1
相减, (n-q)/(m - p) = -4(m+p)/[9(n+q)] (i)
PQ的中点M((m+p)/2, (n+q)/2)
令M(x, y):
m + p = 2x (ii)
n + q = 2y (iii)
PQ的斜率k = (n - q)/(m - p) = -4(m+p)/[9(n+q)]
MA的斜率k' = (y - 1)/(x - 2)
k = k'
(y - 1)/(x - 2) = -4*2x/(9*2y)
4x²- 8x + 9y² - 9y = 0
(3)
a = 4, 按椭圆定义,|F1C| + |F2C| = 2a = 8, |F1D| + |F2D| = 2a = 8
△F2CD的周长 = |F1C| + |F2C| + |F1D| + |F2D| = 16
3+k > 0, k > -3
2 - k > 0, k < 2
-3 < k < 2
最后一题不太清楚。
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1,2x^2-y^2-4x-y=0
2,4x^2+9y^2-8x-9y=0
3,△F2CD的周长为16
2,4x^2+9y^2-8x-9y=0
3,△F2CD的周长为16
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1.点差法
1.
设P(m, n), Q(p, q):
m² - n²/2 = 1
p² - q²/2 = 1
相减, (n-q)/(m - p) = 2(m+p)/(n+q) (i)
PQ的中点M((m+p)/2, (n+q)/2)
令M(x, y):
m + p = 2x (ii)
n + q = 2y (iii)
PQ的斜率k = (n - q)/(m - p)
MA的斜率k' = (y - 1)/(x - 2)
k = k', (x - 1)/(y - 2) = (n - q)/(m - p) = 2(m+p)/(n+q) (i)
= 4x/(2y) (ii)(iii)
= 2x/y
2x² - 4x - y² + y = 0
2.
与1类似
设P(m, n), Q(p, q):m²/9 + n²/4 = 1
p²/9 + q²/4 = 1
相减, (n-q)/(m - p) = -4(m+p)/[9(n+q)] (i)
PQ的中点M((m+p)/2, (n+q)/2)
令M(x, y):
m + p = 2x (ii)
n + q = 2y (iii)
PQ的斜率k = (n - q)/(m - p) = -4(m+p)/[9(n+q)]
MA的斜率k' = (y - 1)/(x - 2)
k = k'
(y - 1)/(x - 2) = -4*2x/(9*2y)
4x²- 8x + 9y² - 9y = 0
3.画图
①△F2CD的周长=4a=16
②3+k>2-k>0⑴即k∈(-1/2,2)
2-k>3+k>0⑵即k∈(-3,-1/2)
综上
③y=kx+b
Y²/9+X²=1
Δ>0
x1+x2=-1/2
1.
设P(m, n), Q(p, q):
m² - n²/2 = 1
p² - q²/2 = 1
相减, (n-q)/(m - p) = 2(m+p)/(n+q) (i)
PQ的中点M((m+p)/2, (n+q)/2)
令M(x, y):
m + p = 2x (ii)
n + q = 2y (iii)
PQ的斜率k = (n - q)/(m - p)
MA的斜率k' = (y - 1)/(x - 2)
k = k', (x - 1)/(y - 2) = (n - q)/(m - p) = 2(m+p)/(n+q) (i)
= 4x/(2y) (ii)(iii)
= 2x/y
2x² - 4x - y² + y = 0
2.
与1类似
设P(m, n), Q(p, q):m²/9 + n²/4 = 1
p²/9 + q²/4 = 1
相减, (n-q)/(m - p) = -4(m+p)/[9(n+q)] (i)
PQ的中点M((m+p)/2, (n+q)/2)
令M(x, y):
m + p = 2x (ii)
n + q = 2y (iii)
PQ的斜率k = (n - q)/(m - p) = -4(m+p)/[9(n+q)]
MA的斜率k' = (y - 1)/(x - 2)
k = k'
(y - 1)/(x - 2) = -4*2x/(9*2y)
4x²- 8x + 9y² - 9y = 0
3.画图
①△F2CD的周长=4a=16
②3+k>2-k>0⑴即k∈(-1/2,2)
2-k>3+k>0⑵即k∈(-3,-1/2)
综上
③y=kx+b
Y²/9+X²=1
Δ>0
x1+x2=-1/2
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设直线为Y-1=k(X-2),PQ中点坐标为(X中,Y中)
中点k=b²×X中/a²×Y中∴k=2X中/Y中
代入得∶2X²-Y²-4X+Y=0
2, 椭圆中过直线中点的斜率为双曲线的相反数
∴4x^2+9y^2-8x-9y=0
3,△F2CD的周长为4a=16
3+k>0,2-k>0,3+k≠2-k求交集就好
呃……求L斜率的取值范围吧,﹙﹣√3/3,0﹚,﹙0,√3/3﹚
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