Page 61 Elective 2 1, Attention in Simple Geometric Properties of Hyperbola, See Supplementary Quest

Solution: Both "real axis length" and "imaginary axis length" have "definitions" in the book.

For hyperbola (x a) y b ) = 1 where a > 0 and b > 0, we call 2a the length of the real axis of this hyperbola, and 2b is called the length of the imaginary axis.

For hyperbola (y a x b ) = 1 where a > 0 and b > 0, we call 2a the real axis length of this hyperbola and 2b the imaginary axis length.

So, if and only if a=b=( 2 2)c (where c is the half-focal length), the real and imaginary axes are equal in length.

Graphically, the dashed axis length is the distance between point a(0,b) and point b(0, b) in the Cartesian coordinate plane where the hyperbola is located.

Take the first hyperbola as an example, let x = 0 in the equation, get y = b, then y has no real solution, and the value of y cannot be calculated. However, when you learn Complex Numbers, you will know that y has an imaginary solution in the complex number set, and you will have the ability to solve the imaginary number solution of y as y=bi, so as to understand more deeply the concepts of "imaginary axis" and "imaginary axis length".

The length of the real axis = 2a, the length of the imaginary axis = 2b, and the length of the real axis = the length of the imaginary axis when there is a real axis length = the length of the imaginary axis, so the hyperbola is called the equiaxed hyperbola.

It's all defined in the textbooks.

In the hyperbolic standard equation, 2a is the length of the real axis, 2b is the length of the imaginary axis, and the distance between the two endpoints in the image is the length of the real axis, and the imaginary axis is invisible.

When the slope of the asymptote is plus or minus 1, it is an equiaxed hyperbola (i.e., a=b).

Solution: Let m(x, y) be any point on the hyperbola, f1 is to the left of f2, and 1) if the point m is on the right branch of the hyperbola, <>

By |mf1|is the length of the line segment mf1, mf2|is the length of the line segment mf2, and the line segment mf1 is longer than the line segment mf2.

mf1|>|mf2|.

Thus |mf1|-|mf2|>0

Go to the absolute value symbol, there is.

mf1|-|mf2||=|mf1|-|mf2|

When the point m is on the right branch of the hyperbola, |mf1|-|mf2|=2a(a>0)

2) The same is true. By |mf1|is the length of the line segment mf1, mf2|is the length of the line segment mf2, if the point m is on the left branch of the hyperbola, as shown below.

The line segment mf1 is shorter than the line segment mf2, |mf1|<|mf2|.

Thus |mf1|-|mf2|<0

Go to the absolute value symbol, there is.

mf1|-|mf2||=-（|mf1|-|mf2|）=-2a.

When the point m is on the left branch of the hyperbola, |mf1|-|mf2|=-2a(a>0)

3) Definition of hyperbola: ||mf1|-|mf2||=2a with the definition of an ellipse: |mf1|+|mf2|=2a is different in two ways:

The former is a constant value of the difference between any point m on the hyperbola and the distance between the two fixed points f1 and f2;

The latter is the sum of the distances from any point m on the ellipse to the two fixed points f1 and f2 and is equal to constants, and one is the absolute value of the difference between the two distances (i.e., the two line segments).

the other is the sum of two distances (i.e., two line segments);

It's really untouchable!

The magnitude of the constant value 2a is different.

hyperbola: 0<2a<2c; (A is smaller than C).

Whereas. Ellipse: 0<2c<2a(A is larger than C).

25x^2-16y^2=400 ===> x^2/16-y^2/25=1

So, a 2 = 16 and b 2 = 25;c 2 = a 2 + b 2 = 41 then a = 4, b = 5, c = 41

The length of the real axis = 2a = 8, the length of the imaginary axis = 2b = 10, the vertex coordinates (4,0), the focal coordinates (41,0), the asymptote equation y = (5 4)x, the eccentricity e = c a = 41 4

Question 2: The length of the real axis = 2a = 8, the length of the imaginary axis = 2b = 4 2 Question 3: The asymptote is y = (1 2)x

Question 4: a=2, b=2 2, c=2 3, so eccentricity e= 3 Question 5: a= a, b= b, c= (a+b) So, e=c a= (a+b) a= 5 2 ===> (a+b) a=5 4 ===> b a=1 4 ===> a=4b

The asymptote is y= (b a)x, and the vertex is (a,0), so the distance from the vertex to the asymptote is d=|√b|/√[(b/a)+1]=2√5/5===> √b/[(1/4)+1]=2√5/5===> √b=(2√5/5)*(5/4)=√5/2===> b=5/4

So, a=4b=5

Then the hyperbolic equation is: x 2 5-y 2 (5 4)=1

Although it is on the y-axis, it is not on the curve, that is, b1 and b2 are special points defined by people to facilitate the study of hyperbola, and this point is not on the curve, so the line segment b1b2 is called the imaginary axis (i.e., the axis that does not exist). The axis was drawn just for the convenience of the study to imitate the definition of the minor axis of the ellipse. If you look at the diagram of the buried hyperbola, it is easy to understand the chain ant.

Have you ever learned imaginary numbers? y 2=-b 2, so y=bi or -bi, i is an imaginary unit and i 2=-1 is defined, so y is a real part of zero, the imaginary part is b or -b pure virtual resistance is all, and y's modulo is |b|, and the imaginary number is similar to a vector, and can be represented in a Cartesian coordinate system, such as y=bi in a Cartesian coordinate system as (0, b).

In the second semester of the second semester of high school, there will be a chapter on imaginary numbers, and you will understand it by then.

The equation does not have a real root, which means that the curve has no intersection with the x-axis. b1(0,-b), b2(0,b) are two points found according to b in the equation, not two points drawn according to the intersection, so they can be drawn and are exactly on the y-axis. This is not a contradiction.

The imaginary axis and imaginary half axis mentioned later are definitions, that is, they are prescribed as such.

For example, the curve is a semicircle on the unit, and the equation f(x,y)=x 2+y 2-1=0 represents the unit circle, so the coordinates on the curve are the solutions of this equation, but the solution of this equation also includes the lower semicircle of the unit, so condition (2) is necessary.

But if the curve is a unit circle, the equation f(x,y)=y-(1-x 2) (1 2)=0

The equation represents the upper half unit circle, so the solution of this equation is on the curve, but the curve also includes the lower half unit circle, so the points on the curve are not all solutions to this equation, so condition (1) is necessary.

Therefore, only when conditions (1) and (2) are satisfied at the same time can the curve and equation be consistent.