High score to solve sophomore math problems, urgent, master advance

The range of k: [-1 3,0].

Idea: After the linear l equation of a and b, we can get: (x a) + (y -b) = 1 (the intercept formula of the linear equation, this will not be solved by other methods to solve the equation l) to obtain: -bx+ay+ab=0

And the distance from the origin to l is: |ab|/√(a^2+b^2)=√3/2………1 a 2 denotes the square of a.

e=c/a=2√3/3………2

It can be solved: a 2 = 3, b 2 = 1The hyperbolic equation is known.

From the point a(0,-1), it is obvious that k must exist, otherwise the straight line is parallel to the y-axis, and it does not conform to c and d on a circle with a as the center. The following is a discussion of the situations in which k exists:

Substituting the line m:y=kx+m into the hyperbola, we get an equation about k,m:

1-3(k 2)] x 2-6kmx-3(m 2)-3=0, x1+x2=6km [1-3(k2)].

Then y=kx+m gives y1+y2=2m+

where (x1,y1)(x2,y2) are the coordinates of points c and d).

Let the midpoint of c and d be n, and the point on the circle with a as the center of c and d is the center, so the slope of points a and n is -1 k

Using the slope formula, we can get: k 2 = -(1+3m) 3m, obviously this formula is greater than or equal to 0, and the range of k can be obtained by solving it: [-1 3,0].

Don't you think it's okay to just have k=0 m as any value?

1.x^2/a^2+y^2/b^2=1

From the crossing point (2,- 2 2) we get 4 a 2+1 (2b 2)=1 and from the crossing point (- 2,- 3 2) we get 2 a 2+3 (4b 2)=1 to get a 2=8

b 2 = 1 so the standard equation is: x 2 8 + y 2 = 1

2.Let the equation be x 2 a 2 + y 2 b 2 = 1 and give c = 2 from the focal length of 4

a^2=b^2+c^2=b^2+4

It is obtained by (3,-2 6) 9 a 2 + 24 b 2 = 1 joint solution.

a^2=36

b^2=32

So the standard equation is x 2 36 + b 2 32 = 13

a^2=16

a=4ab+af2+bf2=af1+bf1+af2+bf2=4a=16 (according to the ellipse definition, af1+af2=bf1+bf2=2a).

4.I can't see the diagram.

1.Establish.

x^2/a^2+y^2/b^2=1

Crossing the point (2,- 2 2) gives 4 a 2+1 (2b 2)=1 passing the point (- 2,- 3 2) gives 2 a 2+3 (4b 2)=1 to get a 2=8

b 2 = 1 so the standard equation is: x 2 8 + y 2 = 1

2.Let the equation be x 2 a 2 + y 2 b 2 = 1 and the focal length is 4 to get c = 2

a^2=b^2+c^2=b^2+4

After (3,-2 6) we get 9 a 2 + 24 b 2 = 1 solution.

a^2=36

b^2=32

The standard equation is x 2 36 + b 2 32 = 1

3.a^2=16

a=4ab+af2+bf2=af1+bf1+af2+bf2=4a=16

4.Draw it and it will come out, and then find two points and bring them into the standard equation to find it OK

x-5)/(x^2-2x-3)>1

x-5)/(x^2-2x-3)-1>0

x-5)-(x 2-2x-3)] (x 2-2x-3) >0-x 2+3x-2) (x 2-2x-3)>0-x 2+3x-2)(x 2-2x-3)>0x 2-3x+2)(x 2-2x-3)<0x-1)(x-2)(x-3)(x+1)<0 to get the shed:

1< x<1,2 "Nai Tan x<3< p>

First move the item, move the 1 to the left;

Then divide and make the denominator of 1 into x flat wheel clear square multiplied by 2x minus 3;

After subtracting the fractions, a fraction will be obtained before the left wax number;

Then formula, draw, and take the part greater than 0.

It's a little stiff after the words.,I don't know if the method is okay.,But I hope it can help you.。

Move items, merge. Get (x-1)*(x-2) >0The numerator and denominator infiltrate the ants and draw quadratic diagrams respectively. The two clusters of the numerator denominator (the positive and negative denominator of the numerator and the negative denominator are positive) are discussed in different cases, and the solution of Fan Hail Nawei (-1,1) and (2,3) is obtained

x-5)/(x^2-2x-3)>1

x-5)/(x^2-2x-3)-1>0

x 2-3x+2) Finger bucket (x 2-2x-3) >0x 2-3x+2) (x 2-2x-3) <0x-1) (x-2) (x+1) (x-3) < 0 draw the number line.

1 "Feng Pai X< 1 or 2 "Wei Ji Mo X< 3

In the first step, the numerator x-5 denominator x square-2x-3 then withers minus 1 greater than 0

In the second step, the numerator x squared minus x plus 2 denominator x squared minus 2x minus 3 is greater than 0

In the third step, the numerator x minus 2 denominator x minus 3 is less than 0 in the fourth step, the rotten spike brace on x minus 2 is greater than 0 and the brace below x minus 3 is less than 0 or the curly brace is hungry.

A 2 = 100 c = 64 can be obtained if there is a title

b^2=a^2-c^2=36

Triangle area s=1 2 b 2 tan angle f1pf2 = 18 3 for triangle area s=1 2 b 2 tan angle f1pf2 you can go and ask your teacher, or push it yourself, it is easier to push it when it coincides with point b with p point.

Take the ab side as the x-axis, then a(-a,0)b(a,0), let c be (x,y) then d is ((a+x) 2,y 2), and according to ad=m, use the distance formula, because the coordinates of a, d points are already known, so you can solve the trajectory equation about xy (a, b two points should be y-axis symmetry).

With A as the 0 point and AB as the X axis to establish the coordinate system Yox, let the coordinates of point C be (X, Y), because AD is the middle line on the edge of BC, then the coordinates of point D are (A+,, there is (A+, solution (X+2A) +Y =(2M), we can see that the trajectory equation of vertex C is centered on the point on the extension line of B with a distance of 2a as the center of the circle, and 2m as the radius of the circle (except for the two points on the same line as AB).

Establish a coordinate system with a as the 0 point and ab as the x-axis

b(2a,0) because AD is the midline on the side of bc, let d (MCOS, MSIN).

C(2MCOS -2A,2MSIN) is obtained from the image(x+2a)2+y2=4m2

With ab as the x-axis to establish the coordinate system, if the coordinates of point c are (x,y), then the d coordinates are (x 2-3a, 2,y 2), then (x 2-3a 2) 2+(y 2) 2=m 2 is the result.

What's wrong with this topic! You can make it!

Slope k== 25-a2) a

The slope is the tangent of the angle.

a<0, the angle is obtuse, so the slope is negative.

Use a to find tangents.

(1) Vector on= vector oa + (1-) vector ob Add vector ao on both sides of the equation, since ao= ao+(1- )ao, we can know that vector an= vector 0+(1-) vector ab Therefore, a, b, n three points are collinear.

2) Since for any of the Interval [0,1], the vector on=root2(1-) is followed by 2

y'=1-1/x²

Because x 0, 1 x 0

So y'1

The derivative of the above equation is based on the derivative formula (x n).'=nx (n-1) where 1 x can be seen as the negative primary square of x.

I hope it can help you, and if you still don't understand, you can hi me.

Just learn derivatives. The question is simple.

The slope of the straight line is the derivative y of the function y'=(x)'+(1/x)'=1-1 x 2, one square of x is greater than 0, the negative number is the opposite, less than zero, 1 plus a negative number must be less than 1, the proof is completed.