数学题目!求高手解答!
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1.设抛物线方程为y^2=2px,
由于直线y=2x+1于抛物线交与两点A,B,带入,得
(2x+1)^2=2px==>
4x^2+2(2-p)x+1=0
△=4(2-p)^2-16=4(4-4p+p^2)-16=4(p^2-4p)=4p(p-4)>0==>p<0或者p>4
√15=
|AB|=√(1+2^2)△/|a|=√[20p(p-4)]
/
4
==>15*16=20(p^2-4p)==>p^2-4p-12=0
==>(p+2)(p-6)=0
==>p=-2
或者p=6
所以抛物线为
y^2=
-4x或者y^2=12x
2.椭圆的焦点坐标为
F1(-c,
0)
,
F2(c,
0),
抛物线的焦点为F(b/2
,
0)
|F1F2|=2c==>|F1F|=2c
*
5/(5+3)=5c/4=c+b/2==>c/4=b/2==>b=2c
e=c/a=c/√(b^2+c^2)=c/√(4c^2+c^2)=1/√5=√5/5.
3.直线y=ax+1与双曲线3x²-y²=1相交于M,N两点,设M(x1,y1),
N(x2,y2)
3x^2-(ax+1)^2=1==>(3-a^2)x^2-2ax-2=0==>△=4a^2+8(3-a^2)=24-4a^2=4(6-a^2)==>-
√6
≤a≤√6
|MN|^2=(1+a^2)|x1-x2|^2
=
(1+a^2)△/(3-a^2)^2
=
4(1+a^2)(6-a^2)/(3-a^2)^2
MN的中点为Q((x1+x2)/2,
(y1+y2)/2)
x1+x2=2a/(3-a^2),
y1+y2=a(x1+x2)+2=2a^2/(3-a^2)+2=6/(3-a^2),
==>Q(a/(3-a^2),
3/(3-a^2),
),
MN为直径的圆过原点,Q为圆心,所以2OQ=MN==>
4OQ^2=MN^2
4{[
a/(3-a^2)
]^2
+[3/(3-a^2)
]^2}=4(1+a^2)(6-a^2)/(3-a^2)^2
(a^2+9)/(3-a^2)^2
=
(6-a^2)(1+a^2)/(3-a^2)^2
a^2+9=(6-a^2)(1+a^2)
a^4-4a^2-3=0
a^2=1,
a^2=-3(舍)
a=±1
4. 圆x^2+y^2=17的圆心为O(0,0),
OA的斜率为
-1/4,所以圆在点A的切线方程为
y=4(x-4)-1=4x-17
所以双曲线的一条渐进线为
y=4x==>
两条渐近线满足方程
y^2
=16
x^2
,即
x^2
-
y^2/16
=0
所以设双曲线的方程为
x^2
-
y^2/16
=
±k
由于A在双曲线上,k
=
±[4^2
-
(-1)^2/16]=
±(16-1/16)=±
255/16
所以,双曲线方程为
16x^2
/
255
-
y^2
/
255=±1
。、
5.将直线带入椭圆方程,mx^2
+n
(1-x)^2=1
==>(m+n)
x^2-2nx
+(n-1)=0,
方程的解就是A,B的横坐标,设A(x1,y1),
B(x2,y2)
x1+x2=2n/(m+n),y1+y2=(1-x1)+(1-x2)=2-(x1+x2)=2-2n/(m+n)=2m/(m+n)
所以AB的中点C的坐标为
(n/(m+n),
m/(m+n))
OC的斜率为
√2/2
=
[n/(m+n)]
/
[m/(m+n)]
=n/m
由于直线y=2x+1于抛物线交与两点A,B,带入,得
(2x+1)^2=2px==>
4x^2+2(2-p)x+1=0
△=4(2-p)^2-16=4(4-4p+p^2)-16=4(p^2-4p)=4p(p-4)>0==>p<0或者p>4
√15=
|AB|=√(1+2^2)△/|a|=√[20p(p-4)]
/
4
==>15*16=20(p^2-4p)==>p^2-4p-12=0
==>(p+2)(p-6)=0
==>p=-2
或者p=6
所以抛物线为
y^2=
-4x或者y^2=12x
2.椭圆的焦点坐标为
F1(-c,
0)
,
F2(c,
0),
抛物线的焦点为F(b/2
,
0)
|F1F2|=2c==>|F1F|=2c
*
5/(5+3)=5c/4=c+b/2==>c/4=b/2==>b=2c
e=c/a=c/√(b^2+c^2)=c/√(4c^2+c^2)=1/√5=√5/5.
3.直线y=ax+1与双曲线3x²-y²=1相交于M,N两点,设M(x1,y1),
N(x2,y2)
3x^2-(ax+1)^2=1==>(3-a^2)x^2-2ax-2=0==>△=4a^2+8(3-a^2)=24-4a^2=4(6-a^2)==>-
√6
≤a≤√6
|MN|^2=(1+a^2)|x1-x2|^2
=
(1+a^2)△/(3-a^2)^2
=
4(1+a^2)(6-a^2)/(3-a^2)^2
MN的中点为Q((x1+x2)/2,
(y1+y2)/2)
x1+x2=2a/(3-a^2),
y1+y2=a(x1+x2)+2=2a^2/(3-a^2)+2=6/(3-a^2),
==>Q(a/(3-a^2),
3/(3-a^2),
),
MN为直径的圆过原点,Q为圆心,所以2OQ=MN==>
4OQ^2=MN^2
4{[
a/(3-a^2)
]^2
+[3/(3-a^2)
]^2}=4(1+a^2)(6-a^2)/(3-a^2)^2
(a^2+9)/(3-a^2)^2
=
(6-a^2)(1+a^2)/(3-a^2)^2
a^2+9=(6-a^2)(1+a^2)
a^4-4a^2-3=0
a^2=1,
a^2=-3(舍)
a=±1
4. 圆x^2+y^2=17的圆心为O(0,0),
OA的斜率为
-1/4,所以圆在点A的切线方程为
y=4(x-4)-1=4x-17
所以双曲线的一条渐进线为
y=4x==>
两条渐近线满足方程
y^2
=16
x^2
,即
x^2
-
y^2/16
=0
所以设双曲线的方程为
x^2
-
y^2/16
=
±k
由于A在双曲线上,k
=
±[4^2
-
(-1)^2/16]=
±(16-1/16)=±
255/16
所以,双曲线方程为
16x^2
/
255
-
y^2
/
255=±1
。、
5.将直线带入椭圆方程,mx^2
+n
(1-x)^2=1
==>(m+n)
x^2-2nx
+(n-1)=0,
方程的解就是A,B的横坐标,设A(x1,y1),
B(x2,y2)
x1+x2=2n/(m+n),y1+y2=(1-x1)+(1-x2)=2-(x1+x2)=2-2n/(m+n)=2m/(m+n)
所以AB的中点C的坐标为
(n/(m+n),
m/(m+n))
OC的斜率为
√2/2
=
[n/(m+n)]
/
[m/(m+n)]
=n/m
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