1. Let a straight line pass through the point (0, a), the slope is 1, and it is tangent to the circle x2 + y2 = 2x, then the value of a is 2. In a pyramid v-abcd, if the bottom surface ABCD is a square with side length 2 and the other four sides are isosceles triangles with side length 5, then the plane angle of the dihedral angle v-ab-c is What are the ways to judge dihedral angle

1. Let a straight line pass through the point (0, a), the slope is 1, and it is tangent to the circle x2 + y2 = 2x, then the value of a is 2. In a pyramid v-abcd, if the bottom surface ABCD is a square with side length 2 and the other four sides are isosceles triangles with side length 5, then the plane angle of the dihedral angle v-ab-c is What are the ways to judge dihedral angle

1. The linear equation is
y=x+a
If it is tangent to a circle
Then there is an intersection point with x ^ 2 + y ^ 2 = 2x
That is, x ^ 2 + (x + a) ^ 2 = 2x
x^2+x^2+2ax+a^2-2x=0
2x^2+(2a-2)x+a^2=0
Discriminant = (2a-2) ^ 2-4 * 2A ^ 2 = 0
4a^2-8a+4-8a^2=0
-4a^2-8a+4=0
a^2-2a+1=0
(a-1)^2=0
a=1
2. Take AB, CD, midpoint e, F, connect VE, VF, EF
Because e is the midpoint of ab
Because f is the midpoint of CD
Because ABCD is a square
So EF is perpendicular to ab
Because VAB is an isosceles triangle
So ve bisects AB vertically
So the angle VEF is the required value
Similarly, VF splits CD vertically
So ve = VF = radical [(radical 5) ^ 2 - (2 / 2) ^ 2] = 2
EF=2
So the triangle EVF is an equilateral triangle
So the angle VEF is 60 degrees
So the dihedral angle v-ab-c is 60 degrees

A linear equation with a slope of 2 and a chord length of 6 cut in the circle (X-2) ^ 2 + (Y-3) ^ 2 = 16

Let the distance between the center (2,3) of the circle (X-2) ^ 2 + (Y-3) ^ 2 = 16 and the straight line be d,
From the vertical diameter theorem and Pythagorean theorem, we get: D & # 178; = 16 - (6 / 2) &# 178; = 7, that is: D = √ 7,
According to the meaning of the question, let the linear equation be y = 2x + B, that is, 2x-y + B = 0,
According to the formula of distance from point to line, the distance d from the center of circle (2,3) to the line 2x-y + B = 0 satisfies:
The solution is: B = - 1 + √ 35 or B = - 1 - √ 35,
So: the linear equation is: 2x-y-1 + √ 35 = 0 or 2x-y-1 - √ 35 = 0

If the chord length of the line L with slope 1 cut by circle x2 + y2 = 4 is 2, then the equation of the line L is______ ．

Let the equation of a straight line be y = x + B, and the distance from the center of a circle to the straight line be d = | B | 2. Then, from the square of the radius equal to the sum of the square of the distance from the center of a circle to the straight line and the square of half the chord length, we get (| B | 2) 2 + 1 = 4, and the solution is b = ± 6, so the answer is y = x ± 6

L with a slope of - 2 is cut by circle x2 + y2 = 4, and the length is 2. The equation of a straight line is obtained

The length of L with a slope of - 2 cut by circle x2 + y2 = 4 is 2. The equation for finding a straight line is l: y = - 2x + m, 2x + y-m = 0o: x2 + y2 = 4, r = 2, (0,0) and the distance between L and squareroot (3) | 0 + 0-m | / squareroot (2 * 2 + 1) = squareroot (3) | m | = squareroot (15) l: y = - 2x + squareroot (15) or y = - 2x -

If the chord length of the line L with slope 1 cut by circle x2 + y2 = 4 is 2, then the equation of the line L is______ ．

Let the equation of a straight line be y = x + B, and the distance from the center of a circle to the straight line be d = | B | 2. Then, from the square of the radius equal to the sum of the square of the distance from the center of a circle to the straight line and the square of half the chord length, we get (| B | 2) 2 + 1 = 4, and the solution is b = ± 6, so the answer is y = x ± 6

The equation of a line with a slope of 34 and a circumference of 12 of a triangle surrounded by two coordinate axes is______ ．

Let the linear equation be y = 34x + B, let x = 0, then y = B; let y = 0, then x = - 43b. | | + | - 43b | + B2 + 16b29 = 12, | | + 43 | + 53 | - B | = 12, | - B = ± 3. The obtained linear equation is y = 34x ± 3, that is, 3x-4y + 12 = 0, or 3x-4y-12 = 0, so the answer is 3x-4y + 12 = 0, or 3

If the focus f passing through the parabola y2 = 2x, the line L with inclination angle π 4 intersects the parabola at a, B (XA > XB), then the value of | AF | BF |___ ．

The focus f (12, 0) of parabola y2 = 2x can be set as a straight line L: y = X-12 and parabola. After sorting out, we can get: x2-3x + 14 = 0, and the solution is: x = 3 ± 222. From the problem, we can get: XA = 3 + 222, XB = 3-222. From the definition of parabola, we can know: | AF | = XA + 12, | BF | = XB + 12 | AF | BF | = 2 + 22-2 = 3 + 22, so the answer is: 3 + 22

If M = {0, 1, 2}, n = {(x, y) | x-2y + 1 ≥ 0 and x-2y-1 ≤ 0, x, y ∈ m}, then the number of elements in n is______ ．

Draw the feasible region represented by the set n = {(x, y) | x-2y + 1 ≥ 0 and x-2y-1 ≤ 0, x, y ∈ m}, as shown in the figure. From the meaning of the question, we can see that there are only four points (0,0), (1,0), (1,1) and (2,1) in n that satisfy the condition, so the answer is: 4

Through the fixed point (- 2, - 4), make a straight line L intersecting parabola with an inclination angle of 45 °, y2 = 2px and B, C two points, when AB, BC, AC are in equal proportion sequence, find the equation of parabola
Through the fixed point a (- 2, - 4), make a straight line L intersecting parabola with an inclination angle of 45 ° y2 = 2px and B, C. when AB, BC, AC are in equal proportion sequence, the equation of parabola is obtained

Let B (x1, Y1) C (X2, Y2)
Make a straight line L with an inclination angle of 45 ° through the fixed point (- 2, - 4)
Then the linear equation is y = X-2 substituted by y2 = 2px
x^2-(2p+4)x+4=0
x1+x2=2p+4
x1*x2=4
AB, BC and AC are equal proportion series
Then AB / BC = BC / AC
(x1+2)/(x2-x1)=(x2-x1)/(x2+2)
Well organized
x1x2+2(x1+x2)+4=(x1+x2)^2-4x1x2
4+2(2p+4)+4=(2p+4)^2-16
The solution is p = 1
So the parabolic equation is
y^2=2x

If M = {0, 1, 2}, n = {(x, y) | x-2y + 1 ≥ 0 and x-2y-1 ≤ 0, x, y ∈ m}, then the number of elements in n is______ ．

Draw the feasible region represented by the set n = {(x, y) | x-2y + 1 ≥ 0 and x-2y-1 ≤ 0, x, y ∈ m}, as shown in the figure. From the meaning of the question, we can see that there are only four points (0,0), (1,0), (1,1) and (2,1) in n that satisfy the condition, so the answer is: 4