Mathematical hyperbola, questions about mathematical hyperbola

1) If the left vertex is (-1,0), then a=1, and e=root3=c a, so c=root3, so b2=c2-a2=2

So the hyperbolic equation is x 2-y 2 2 = 1

2) Substitute x=y+m(*) into x2-y 2 2=1 and simplify x2-2mx-m 2-2=0

Then x1+x2=2m, x1 x2=-m 2-2, substituting (*) gives y1+y2=x1+m+x2+m=4m, y1 y2=(x1+m)(x2+m)=2m 2-2

(x1,y1) and (x2,y2) are the coordinates of points A and B.

So where the point coordinates are x0=(x1+x2) 2=m, y0=(y1+y2) 2=2m

So, m2+(2m) 2=5, m=1 or m=-1.

When m=1, x1+x2=2,x1 x2=-3,(x1-x2) 2=(x1+x2) 2-4x1 x2=16,y1+y2=4,y1 y2=0,(y1-y2) 2=(y1+y2) 2-4y1 y2=16, so |ab|2=(x1-x2) 2+(y1-y2) 2=32, so |ab|= root number (32) = 4 root number (2).

When m=1, x1+x2=2,x1 x2=-3,(x1-x2) 2=(x1+x2) 2-4x1 x2=16,y1+y2=4,y1 y2=0,(y1-y2) 2=(y1+y2) 2-4y1 y2=16, so |ab|2=(x1-x2) 2+(y1-y2) 2=32, so |ab|= root number (32) = 4 root number (2).

Since c a = 3, i.e. c = 3a, so f( 3a,0) and b =c -a =2a , substituting the straight line y=x-1 into the hyperbola x a -y (2a)=1, we get x +2x-1-2a =0, let a(x1,y1), b(x2,y2).

Then x1+x2=-2, x1x2=-1-2a, y1y2=(x1-1)(x2-1)=x1x2-(x1+x2)+1=2-2a, because the vector fa* vector fb=(x1- 3a)(x2- 3a)+y1y2=4, that is, a-2 3a+3=0, so a= 3, so the hyperbolic equation is x3-y 6=1.

Solution: Let p(x,y).

Since :p is on the right branch of the hyperbola.

Then there is: x>=a

Again: (pf1) pf2=8a

The focal radius formula is used: PF1=EX+A, PF2=EX-A(E=C A).

Substituting (2) obtains:

ex+a)^2/(ex-a)=8a

ex+a)^2=8a(ex-a)

c/a)^2x^2+2cx+a^2=8cx-8a^2(c/a)^2x^2-6cx+9a^2=0c^2x^2-6a^2cx+9a^4=0

cx-3a^2)^2=0

Then: x=3a2c

then substitute (3) into (1) to obtain:

3a^2/c>=a

3a/c>=1

3a>=c

Then: 3>=c a

i.e.: e=c a<=3

Again: e is the eccentricity of the hyperbola should be 1

Then: e can be (1,3).

Let p(x,y)f1(-c,0) f2(c,0) is known by definition|pf1|-|pf2|=2a

again pf1 =3 pf2|

So |pf2|=a

(x-c)2+y2=a2

(x-c)2+b2(x2 a2 -1)=a2=>c2x2 a2-2cx+c2=a2+b2=c2, so c2x2 a2-2cx=0

And because x≠0Solution. x=2a2/c

From 2a2 c a, the solution is: e=c a 2

So the range of e is (1,2].

Let f1 and f2 be x and y, respectively

By the cosine theorem: x + y -4 (a + b) = 2xy 1 2.

By hyperbola: x-y=2a.

By the formula for the area of the triangle: x y 3 2 1 2 2 3.

Eliminate x,y in the form to obtain b = 2, and then by the hyperbolic centrifugal chamber or rate of collapse 2 to obtain a = 2 3

So the hyperbolic equation is: x 2 3 -y 2 = 1 It's only been a summer vacation, and I forgot about it, and it took a lot of effort, hehe.

Since such a p exists, there can be four points that satisfy the condition, and the distance from the point p to the x-axis can be found. That is, to find the absolute value of the ordinate of the point p.

The focal coordinates f1(-5,0),f2(5,0), let p(x,y) be made with vectors, so there are vectors f1p(x+5,y), f2p(x-5,y).

Since pf1 pf2, the product of the two vectors is 0, which is reduced to x 2-25 + y 2 = 0

p is also a point on the hyperbola, so there are two equations (x 2) 9-(y 2) 16 = 1 to solve the absolute value of y, and the absolute value of y is equal to 16 5

f1 and f2 are the left and right focal points, respectively.

Let f1p=x1 f2p=x2

x1 - x2 =2a=6

x1^2 + x2^2 = 100

x1 x x2 = 32

Let the distance from the point p to the x-axis be h

h/x1 = x2/2a

h = 16/5

Got it? Of course, you can also use the vector method to do it, but it's more difficult to calculate, and there are some theorems to memorize.

Because it's the right focus, it's obvious that you know.

c = t, and there is (c b 2 a) (c -b 2 a) corresponding to c < = b 2 a

and there is c 2 = a 2 + b 2;

Eliminate an unknown number of substitutions.

Finally, we can find the range of a e.

In general, hyperbola (literally "exceeding" or "exceeding" in Greek) is a type of conic curve defined as two halves of a plane intersecting a straight conical surface.

It can also be defined as the trajectory of a point where the difference in distance from two fixed points (called focal points) is constant. This fixed distance difference is twice as much as a, where a is the distance from the center of the hyperbola to the vertex of the nearest branch of the hyperbola. a is also called the real semi-axis of the hyperbola.

The focal points are located on the through axis, and their middle point is called the center, which is generally located at the origin.

In mathematics, a hyperbola (multiple hyperbola or hyperbola) is a type of smooth curve that lies in a plane and is defined by an equation of its geometric properties or a combination of its solutions. Hyperbolas have two pieces called connected components or branches, which are mirror images of each other, similar to two infinity bows.

A hyperbola is one of three conical sections formed by the intersection of a plane and a bipyramidal. (The other conic parts are parabolas and ellipses, and circles are special cases of ellipses) If the plane intersects the two halves of the cone, but does not pass through the vertices of the cone, the conic curve is hyperbola.

Note: Contrary to the outer quasi-circle, the condition for having an inner quasi-circle is a, so there can only be one of the inner and outer quasi-circles of the hyperbola. In particular, the equiaxed hyperbola (also known as the right-angled hyperbola, satisfying a=b) has neither an inner nor an outer collimation circle.

This property can be simply remembered as follows: the string of any tangent of the quasi-circle in the hyperbola is truncated by the hyperbola, and the opening angle to the center o is a right angle.

m=12, it can be seen from the inscription of the ode to the branches, a 2=4, a=, so the base c = 4. In the hyperbolic digging line, c 2 = a 2 + b 2. From a = 2 and c = 4 to get b 2 = 12. So m=b 2=12

A is a circle with the origin as the center of the circle. I'll give you a picture later. You wait.

Extend f1p and qf2 and intersect m. Since PQ is an angular divination line and is a high line of F1m, QF1M is an isosceles triangle.

f1q＝mq＝qf2＋f2m

qf2 f1q f2q 2a (hyperbolic definition) o, p are both midpoints, so op is the median line of the triangle f1f2m, so the length is a fixed value