Let the two vertices of the equiaxed hyperbola x 2 y 2 a 2 a 0 be A and B

Proof is that the coordinates of a and b are a(-a,0), b(a,0) respectively, and the coordinates of p are (s,t); Then the absolute value of pq=t, the absolute value of qb=(s-a), the absolute value of qa=(s+a), because the absolute value of qb*qa=(s-a)(s+a) = the absolute value of (s2-a2).

Because t 2>a 2, qb*qa=s 2-a 2, and because pq 2=t 2

Because (t,s) is on the hyperbola, so S 2-T 2=A 2, i.e. S 2-A 2=T 2, i.e. qb*qa=pq 2, i.e. qb pq=pq qa, and because pqa= bqp=90 degrees, the triangle pqa is similar to the triangle bqp, so pbq= qpa, so the angle paq+ pbq= paq+ qpa=90 degrees.

Let p(x0,yo), a(-a,0), b(a,0).

So tan paq=|pq|/|aq|=|y0|/|(x0+a)|,tan∠pbq=|pq|/|bq|=|y0|/|(x0-a)|, and because x0 2-y0 2=a 2, |(x0-a)|(x0+a)|=|y0||y0|,y0|/|(x0-a)| y0|/|(x0+a)|=1, so tan paq*tan pbq=1,0< paq,0< pbq, so the two corners are redundant, so: paq+ pbq=90°.

Hyperbola: x 2 a 2-y 2 b 2 = 1 (a>0, b>0).

The real axis 2a=4 3 => a=2 3

The asymptote y= b a*x, i.e. bx-ay=0

focal points f1(-c,0), f2(c,0), and the distance from the focus to the asymptote is d=|bc|/√(a^2+b^2)=bc/c=b=√3

c= (a 2+b 2) = 15, the hyperbolic equation is: x 2 12-y 2 3 = 1

Substituting the straight line y= 3 3*x-2 into the hyperbola, we get x 2 12-( 3 3*x-2) 2 3=1

Arrangement, can be solved on the right branch (x>0), the two intersection points are m(2 3,0), n (14 3,12).

Vector om = (2 3,0), vector on = (14 3,12), vector om + vector on = (16 3,12) = t vector od

Vector od = (16 3 t, 12 t).

and point d on the hyperbola, (16 3 t) 2 12-(12 t) 2 3=1

The solution yields t= 4

The coordinates of the point d on the right branch are (4 3,3) (take t=4).

The distance of the hyperbola is the distance from the focus to the asymptote is b, then b = 3, and 2a = 4 3, then a = 2 3, and the equation is: x 12 y 3 = 1

Solution: (1) From the problem, the difference is a1(-2,0),a2(2,0), let p(x0,y0), q(x0,-y0), then a1p=(x0+2,y0), a2q=(x0-2,-y0).

From a1p a2q=1, we get x20-y20=3 ....①

and p(x0, vitas y0) on the hyperbola, then x202-y20=1 ....②

Synthesis, solution x0 = 2

By the title, x0 0, x0=2

The coordinates of the point t are (2,0).

2) Let the coordinates of the intersection point M of the line A1P and the line A2Q be (X,Y).

From a1, p, m three points collinear, (x0+2)y=y0(x+2) ....

From a2, q, m three points collinear, (x0-2)y=-y0(x-2) ....

Coupling, solution x0 = 2x, y0 = 2yx

p(x0,y0) on hyperbola, (2x)22-(2yx)2=1

The equation for trajectory e is x22+y2=1(x≠0,y≠0).

3) Since the slope of the straight line l is not 0, the draft disturbance of the straight line l can be set as x=ky+1 and substituted into x22+y2=1 to obtain (k2+2)y2+2ky-1=0

Let a(x1,y1),b(x2,y2), then from the relationship between the root and the coefficient, we get y1+y2=-2kk2+2 ....⑤y1y2=-1k2+2 …⑥

fa= fb, there is y1y2= (0).

Divide the square of the formula by to get y1y2+y2y1+2=--4k2k2+2, i.e. +1 +2=--4k2k2+2

From [2,-1], we get +1 +2 0

4k2k2+2≤0，∴0≤k2≤27

ta+tb=（x1+x2-4，y1+y2）

ta+tb|2=（x1+x2-4）2+（y1+y2）2=16-28k2+2+8(k2+2)2

Let t = 1k2+2, 0 k2 27, 716 1k2+2 12, i.e. t [716,12].

ta+tb|2=f（t）=8t2-28t+16=8（t-74）2-172

and t [716,12], f(t) [4,16932].

ta+tb|∈[2，1328]．

Hyperbola: x 2 a 2-y 2 b 2=1(a>0,b>0) real axis 2a=4 3 gives a=2 3

Asymptote y= b a*x, i.e. bx-ay=0 focal points f1(-c,0), f2(c,0).

The focal point to asymptote distance is d=|bc|/√(a^2+b^2)=bc/c=b=√3

Then c= (a2+b2)= 15, so that the hyperbolic equation is: x 2 12-y 2 3=1

Equiaxed hyperbola x 2-y 2 = a2

With the straight line y=ax(a>0) there is no common point, substitute , (1-a 2)x 2=a 2, there is no real root<==>1-a 2<=0,==>a<=-1 or a>=1, is what is sought.

Let p1(x0,y0), p1(x0,-y0), x 2 4-y 2=1a1(2,0)a2(-2,0).

The equation of the straight line A1P1 is y=y0 x0-2(x-2), the equation of the straight line A2P2 is y=y0 x0+2(x+2), and the simultaneous line A1P1,A2P2 obtains the intersection coordinates of (4 x0, 2y0 x0), and p1(x0,y0) on the hyperbola.

x0 4-y0 =1, set the intersection point (x,y),p1(4 x,2y0 2),4 x -y x0 4,and then substitute x0=4 x to get the trajectory equation x 4+y before the trouble=1o( o, I hope the remorse will help you.

From the meaning of the title, we can know the coordinates of point A (-m, -n).

Bring in the equation of a - n = 1 2m

Bring point b to y=x+8

There is -n=m+8

The solution yields m+n=(-8).

m*n=(-1/2)

So 1 n+1 m=(m+n) mn=16

Big brother, I haven't done a question for several years, and now I don't have paper and pen, please ask others, my current level is not as good as yours.

Let p(m,n),q(x,y).

Hyperbola m:xay

The two vertices of the real axis of b1(a0,b0) are a,b,a(-a,0),b(a,0).

qa（-x-a，-y），pa

m-a，-n）

qa pa, (x-a)(-m-a)+ny=0, we get m+a=-nyx+a

In the same way, according to qb pb, m-a=-nyx?a, m2a2n can be obtained

The point p(m,n) of yxa is a moving point on the hyperbolic m except a and b, manb

Sorted out to n2bam2

a2, substitution simplification xay

AB Selection: C

The coordinates of the symmetry and a of the two points with respect to the parallel collision of the y-axis are (a, b) and b(-a, b), because the point a is on the hyperbola y=1 2x, and b is on the straight line y=-x+3, so b=1 2a, b=a+3, so the original formula = (a square + b square) (ab) = 2 (a square + b square) = 2 (a square + a square + 6a + 9) = 2 (2a square + 6a + 9). From the absolute annihilation b = 1 2a, b = a + 3, eliminate b, 2a square + 6a = 1, so the original argument = 2 (1 + 9) = 20