1. Construct a system of quadratic equations of two variables by image method, and make its corresponding linear function pass through (- 3, - 2), (- 1,6) two points 2. Given that the image intersection of the first-order function y = 2x-a and y = 3x + B and a point outside the origin on the x-axis, what is a / (a + b) equal to?

1. Construct a system of quadratic equations of two variables by image method, and make its corresponding linear function pass through (- 3, - 2), (- 1,6) two points 2. Given that the image intersection of the first-order function y = 2x-a and y = 3x + B and a point outside the origin on the x-axis, what is a / (a + b) equal to?

-3x-2y=a
(1) - x + 6y = a, where a can take any value, so that a binary linear equation system is constructed (different values of a can get different equations system)
(2) Because the first-order function intersects the X axis, y = 0, and because the intersection is a point outside the origin, X is not equal to 0
=>a=2x b=-3x =>a/（a+b）=-2

In this paper, the two equations of the system of linear equations of two variables are transformed into the form of linear function, and the image of the function is drawn in the same coordinate system
The solution of the equations can be written out

The intersection is the solution of two equations

How to approximate the solution of the system of linear equations of two variables with the graph of linear function?

Let y = this equation, and then draw an image. The focus of the x-axis intersection is the solution. Our teacher said that the detailed explanation with the diagram is just an empty sleeve. We really need to calculate on the draft paper, and then write the coordinates on the image

Let F1 and F2 be the two focuses of hyperbola x2 / a2-y2 / B2 = 1 (a > 0, b > 0), if F1, F2, P (0, 2b) are the three focuses of an equilateral triangle
Then E =?
be deeply grateful

∵ F1, F2, P (0,2b) are three parts of an equilateral triangle
Ψ F1F2 is the edge of an equilateral triangle and OP is the height of an equilateral triangle
∴2c*√3/2=2b
∴√3c=2b
Square on both sides:
3c²=4b²=4(c²-a²)
∴c²=4a²
∴c²/a²=4
That is E & # 178; = 4, e = 2

It is known that the left and right focal points of the hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 are F1 and F2 respectively, and point P is on the right branch of the hyperbola
And | Pf1 | = 4 | PF2 |, then the maximum value of eccentricity of this hyperbola is?

According to | Pf1 | - | PF2 | = 2A, and | Pf1 | = 4 | PF2 |, so | Pf1 | = 2A / 3, | PF2 | = 8A / 3, then 2C

Given that P is a point on the right branch of the hyperbola x square / a square - y square / b square (a > 0, b > 0), F1 and F2 are the left and right focal points of the hyperbola respectively, and the focal length is 2c, then the abscissa of the center of the inscribed circle of △ pf1f2 is

Given that P is a point on the right branch of the hyperbola x square / a square - y square / b square (A & gt; 0, B & gt; 0), f ∵ P is on the hyperbola, F1 and F2 are the left and right focal points of the hyperbola respectively

As shown in the figure, the straight line y = MX and the hyperbola y = x intersect two points a and B, passing through point a as am, perpendicular to the x-axis, and the perpendicular foot as m, connecting the two points····
As shown in the figure, the straight line y = MX and hyperbola y = x intersect two points a and B, passing through point a as am, perpendicular to X axis, perpendicular to m, connecting BM. If s triangle = 2, then the value of K is (2)

Let a (x, y) then B (- x, - y)
S=0.5*OM*|y-（-y）|=2
=0.5*x*2y=2
k=x*y=2

As shown in the figure, the straight line y = - x + 6 and the hyperbola y = - 1 / X (x)

Substituting y = - x + 6 into y = - 1 / x, there is - x + 6 = - 1 / x-x ^ 2 + 6x = - 1 x ^ 2-6x-1 = 0. Let a coordinate be (x1, Y1), then OA ^ 2 = X1 ^ 2 + Y1 ^ 2 = (6X1 + 1) + (- X1 + 6) ^ 2 = 6X1 + 1 + x ^ 2-12x + 36 = x ^ 2-6x1 + 37 = 1 + 37 = 38. Let y = 0, then x = 6, that is, B coordinate is (6

If a straight line y = 2x and a hyperbola y = 3 / X intersect a (x1, Y1) and B (X2, Y2), then the relationship between OA and ob is______ X1+X2+y1+y2=__________

Let y = 2x and y = x / 3 stand side by side, that is, 2x = x / 3, get x = + - 2 / 2 root sign 6
And then you get the coordinates of A. B, and then you use the distance formula between two points to get the size relationship between OA and ob. Because o is the origin, it's much more convenient
Get y = + - root 3, OA = ob = 3 / 2 root 2 (hope you can understand)
The second empty is 0

As shown in Figure 1, point a is on the positive half axis of the y-axis, and the equilateral triangle AOC is made with OA as the edge
(1) Point B is a moving point on the positive half axis of x-axis, as shown in Figure 1. When point B moves to the position of point D, connect ad. please determine point E in the first quadrant so that △ ade is an equilateral triangle. (2) under the condition of (1), does the size of ∠ ace change during the movement of point B? (3) as shown in Figure 2, rotate the positive △ ade you made in (1) anticlockwise around point a, so that point e falls on the position of E 'on the positive half axis of Y axis, and get the positive △ AE'd', connecting CE 'and od' to point F. now we give two conclusions: ① AF bisection ∠ CAD '; ② FA bisection ∠ ofe', of which there is and only one conclusion Correct, please judge which conclusion is correct and prove it

(1) As shown in the following figure: take a and D as the center of the circle, ad as the radius, draw an arc, take the intersection E in the first quadrant, connect AE and De, then the triangle ade is the equilateral triangle. (2) the size of ∠ ace does not change, and is always equal to 90 ° reason: according to the meaning of the title, there are ad = AE, Ao = AC, ∠ oad + ∠ CAD = ∠ CAE + ∠ CAD = 60 ° and ∠ oad = ∠ CAE, in △ ace and △ AOD, AE = ad ∠ EAC (3) the second conclusion is that it is correct for FA to divide ∠ ofe ′, the reason is: am ⊥ OD ′ in M, an ⊥ CE ′ in N, AE ′ = ad ⊥ e ′ AC = D ′ Aoao = AC, ≌ OAD ≌ CAE ′ (SAS) and ⊥ C in △ OAD ′ and △ CAE ′ E ′ = OD ′, ∧ am = an (the heights of the corresponding sides of congruent triangles are equal), ∧ an ⊥ CE ′, am ⊥ OD ′, ∧ AFN = ∠ AFM, i.e., FA bisects ∠ ofe, ∧ ② is correct; ∧ Fe and of are not equal, ∧ FAE is not necessarily equal to ∠ FAO, ∧ ead ′ = ∠ Cao = 60 °, ∧ D ′ AF is not necessarily equal to ∠ fac, ∧ ① is wrong; that is, only ② is correct