Compulsory 2 Mathematics Analytic Geometry

The circle intersects the line m -4 3 0,+

The circle is tangent to the straight line m=0 or -4 3

The circle is separated from the straight line m -4 3,0

Solution]: The equation for a circle x y -4x-2y + 1 = 0 can be reduced to the standard formula (x-2) +y-1) =2

The center of the circle is c(2,1) and the radius is 2

As long as the distance from the center of the circle c(2,1) to the straight line mx-y-m-1=0 and the magnitude relationship of radius 2 are determined, the position relationship between the line and the circle can be clarified, so that the problem of several common points can be known, and finally m can be solved.

We know that the equation for the line l is ax+by+c=0, the coordinates of the point p are (xo,yo), then the distance from the point p to the line l is:

d=│axo+byo+c│ / √（a²+b²）。

Then the distance from c(2,1) to the straight line mx-y-m-1=0.

d=│m*2+(-1)*1+(-m-1)│ / √[m²+(1)²]=|m-2|/√(m²+1)

1) When a circle has 2 common points with a straight line: d=|m-2|(m +1) 2 solution m (-4 3) (0,+.)

2) When a circle has 1 common point with a straight line: d=|m-2|(m +1)=2 The solution gives m=0 or m= -4 3

3) When there is no common point between the circle and the straight line: d=|m-2|(m +1) 2 solution m (-4 3) (0,+.)

In a system of continuous equations, there are several solutions that have several common points.

First, write the equation for a circle in the form of (x-a)*2+(y-b)*2=r*2. Find the center of the circle (a, b), and then use the formula from point to line, according to 1, 2, 3 to ask different inequalities, and solve it.

Indicates that the cube is connected to the ball. The diagonal of the body of the positive collapse cube is the diameter of the ball.

The diagonal of the cube is 3*acm, so the radius of the ball is r= 3 2*acm

The volume of the sphere is 4 r 3 3 = 4 3*( 3 2*a) 3cm 3= 3 2*a 3cm 3

9 Dust orange 2 a3

The sphere is the center of the cube.

From a vertex of the cube to the center of the cube is the radius of the sphere, and then a perpendicular line of an edge is made through the center to form a right-angled triangle with a vertex.

This perpendicular line is a half-and-a-half of the diagonal length on the cube.

The Pythagorean theorem finds the Qiji radius of the ball.

Then use the formula for the volume of the ball to find the volume of the ball.

Because the vertices of the cube are limb on the spherical surface.

The trembling line is the diagonal of the cube body as the diameter of the ball.

According to the cuboid diagonal formula l 2 = a 2 + b 2 + c to get the root number three of the ball with a diameter of a fold.

The root number three with a radius of one-half a times.

So the volume of the ball is.

The root of the second is number three

The diagonal line of the cube will pass through the center of the ball, so only the diagonal line of the cube can be long.

This should be easy to ask for.,A 2+A 2+A 2=3A 2,Open a Cong letter root number,It's (root 3)a,The volume of the ball is 4 r 3 3,It's inconvenient to play here.,Ask for it yourself.。。。

<1> because AC+AD=2A and CD=2 3

So it is an ellipse: the major semi-axis is a, the minor semi-axis is (a 2-3) the curve equation is: x 2 a 2+y 2 (a 2-3)=1<2> there is a geometric relationship that has a maximum angle cad at the intersection of the y-axis and the curve, so bc so 3(1).

The owner of the watchtower will solve it himself, and he will be more impressed in the future and improve his academic performance!

c = 3, e=c a= 3 2, a=2, b 2=a 2-c 2=1, the elliptic equation is: x 2 4 + y 2 = 1, the coordinates of the upper vertex are (0, 1), with b as the vertex, respectively, make a straight line at an angle of 45 degrees to the y axis on the left and right of the longitudinal axis, and cross the ellipse at two points a and c, because the ellipse is an axisymmetric figure, and the isosceles right triangle is also an axisymmetric figure, so an isosceles right triangle can be made, the axis of symmetry is the y axis, and the b point is the right vertex, and two isosceles right triangles can also be made, The right-angled vertices are not at point b, and the left and right sides are symmetrical.

Therefore, it can be made into three isosceles right triangles.

There are two different intersection points between the quadratic curve and the line segment, which can be obtained by drawing m>0---1), the opening is downward, and the linear equation where the line segment is located is x+y=3

The two equations are -x +mx-1=3-x, i.e., x -(m+1)x+1=0(m+1) -4>0

m>1,m<-3---2) because there are two intersections, so the intersection of the quadratic curve and the right side of the equation is (3,0), so m=10 3, because there are two intersections, so m<10 3

In summary, 1 can be obtained

The line segment equation for a(0,3),b(3,0) is the endpoint: y=-x+3(0<=x<=3).

There are two different intersections.

y=-x²+mx-1

y=-x+3

0<=x<=3)

There are two sets of solutions to these two equations.

i.e. -x+3=-x+mx-1 has two solutions.

x - (1+m) x + 4 = 0

In this range.

0<=x<=3)

There are two solutions on it.

Let f(x)=x -(1+m)x+4

To make f(x)=0 in the interval.

0<=x<=3)

There are two solutions on it.

Then f(x)=0 root, x1>=3, x2<=3

and δ>0 to solve the range of m that can be obtained.

Because the equation for the line segment ab is y=-x+3

0 x 3), substitute y=-x +mx-1 to obtain: x + (1-m) x+4 = 0, the constructor f(x) = x + (1-m) x+4, so f(0)=4 0, f(3)=9+3(1-m)+4=16-3m 0, =(1-m) -16 0, m-1 0, solution: 5 m 16 3

This is done with interpretive geometry, the PQ highway can be regarded as an ellipse focusing on two points b and c, list its equations, and then find the distance from the point on the ellipse to a, it is obvious that the intersection point of the ac line and the ellipse is the m point, and the point will stop, and it should be calculated.

Solution: Taking the straight line where cd is located as the x-axis and the midpoint o of cd as the origin, establish a Cartesian coordinate system, then c(- 3, 0), d( 3, 0).

1) Let a(x,y), from ac+ad=2a, to obtain [(x- 3) 2 +y 2] x+ 3) 2 +y 2] =2a, simplify, get:

When a< 3, it is meaningless; When a = 3, the equation is y=0 (- 3 x 3) and the trajectory is the line segment cd;

When a 3, (a 2-3) x 2 + a 2 y 2 = a 4 -3a 2

2) In an ellipse, when A is the vertex of the minor axis, CAD is maximum. When cad= 2, a 2+a 2=(2 3) 2 a = 6So when 3 A 6, there is point A, making AC perpendicular to AD.

3) When a=2, the moving point b satisfies x +4y =4. (1) Let the straight line of ab be: y=kx+m.......1）

x^2/4+y^2=1...2)

The solution of the two formulas is: x1+x2= ? x1x2=?

y1+y2=? y1y2=?

From the AO vertical ob x1x2 + y1y2 = 0

The distance from the origin o to the straight line d=|m|/√（1+k^2）

ab|=|x1-x2|√（k^2+1）

Area s=1 2d|ab|

2) When the slope k of the straight line ab does not exist (the line is perpendicular to the x-axis), y=m

The solution is the same as above< 3>(1).

The maximum value of the triangle AOB area can be obtained is 1, and the minimum value of the triangle AOB area is 4 5.

Answer: y=2x+2 is a symmetrical straight line about y=x.

Synoptic: y=2x+2=x

Solution: x=-2

The intersection point is (-2, -2) is also on a symmetrical line.

The point (0,2) is on y=2x+2.

Point (0,2) The symmetry point with respect to y=x is (2,0), and the symmetrical straight line passes through the points (-2,-2) and (2,0).

Substituting y=kx+b to get :

y(-2)=-2k+b=-2

y(2)=2k+b=0

Solution: b=-1, k=

The line of symmetry is: y=

Classmate, don't be naughty, you've got me.

Center of gravity, abscissa: (x1+x2+x3) 3 ordinate: (y1+y2+y3) 3

ab=ob-oa

Center = ab 2

I'm too lazy to write, you do the math yourself, you have to memorize the formula.

k = 1 so k'=-1 y-2=-(x+1) y'=-x+1 simultaneous equation where two lines intersect with x=1 . So another point (3,-2).

1.Midpoint m(3,2),a(4,-1).

Then the coordinates of b are (2,5).

Let c(x,y).

x+3+2）/3=4

y-1+5)/3=2

Solution: c(6,2).

bc = 4 + 3 = 5

1) Let b(a,b),c(c,d).

From the midpoint formula, we get a 3 2 4 2

b＝2×2－〔－1〕＝5

b（2，5）

From the formula of the center of gravity, c 4 3 4 2 6 is obtained

d＝2×3－〔－1＋5〕＝2

c（6，2）

From the point distance formula, bc [a c b d ]52) let the straight line through p(-1,2) perpendicular to the straight line y=x 1 as the analytical formula y -x b

Substituting p(-1,2) gives y -x 1

The intersection of two straight lines is the coordinate of k (1,0) and k is p and its symmetry point p'The midpoint of the line.

Let p'(a,b), a 1 2 1 3b 0 2 2 2 2 from the midpoint formula

p'（3，－2）

The diameter of the ball is the body diagonal of the cube.

So the sphere radius r= 3a 2

Then use the ball volume formula.

The center of the diagonal face of the square is the center of the sphere, so the ball r m* 2

v=4/3 *∏m*√2)^3=8 √2 /3*∏m^3

The diagonal of the cube is 3a

The radius of the ball is 3a 2

The volume of the ball = 4 3 pie r 3 = 3 pie a 3 2

The length of the cube to the apex angle is the diameter of the ball: (root number 3) a so, the radius of the sphere (root number 3) a 2

So, the cube of the sphere volume 4 3*{[root number 3)a 2]}* is 9 8*(root number 3)*a*a*a*

The root number can't be typed.。。。