It is known that the ellipse X & # 178; / A & # 178; + Y & # 178; / B & # 178; = 1 (a ＞ 0b ＞ 0) and the hyperbola X & # 178; - Y & # 178; / 2 = 1 have the same focus. F1. F2. P is a common intersection of the hyperbola and the ellipse, and | Pf1 |. | PF2 | = 3 - the equation for finding the ellipse, if the straight line y = x + m intersects the ellipse and a, B, two points, O are the coordinate origin, try to find the maximum area of the triangle AOB

It is known that the ellipse X & # 178; / A & # 178; + Y & # 178; / B & # 178; = 1 (a ＞ 0b ＞ 0) and the hyperbola X & # 178; - Y & # 178; / 2 = 1 have the same focus. F1. F2. P is a common intersection of the hyperbola and the ellipse, and | Pf1 |. | PF2 | = 3 - the equation for finding the ellipse, if the straight line y = x + m intersects the ellipse and a, B, two points, O are the coordinate origin, try to find the maximum area of the triangle AOB

c^2=a^2-b^2 =1+2=3
PF1+PF2=2a lPF1-PF2l=2 PF1*PF2=3 (PF1>PF2)
(PF1+PF2)^2=(PF1-PF2)^2+4PF1*PF2=4+12=16
PF1+PF2=4=2a
a=2
a^2-b^3=3 b=1
So x ^ 2 / 4 + y ^ 2 = 1

3sinx = cosx-1, then the value of TaNx / 2 is

3sinx=cosx-1
6sin(x/2)cos(x/2)=-2[sin(2/x)]^2
sin(x/2)[3cos(x/2)+sin(x/2)]=0
So sin (x / 2) = 0, 3cos (x / 2) + sin (x / 2) = 0
So TaNx / 2 = 0 or - 3

tanx=-1/2,sin²x+3sinxcosx-1=?

sinx/cosx=tanx=-1/2
cosx=-2sinx
So cos & # 178; X = 4sin & # 178; X
Substituting the identity Sin & # 178; X + cos & # 178; X = 1
So Sin & # 178; X = 1 / 5
sinxcosx
=sinx(-2sinx)
=-2sin²x
=-2/5
So the original formula = - 2

If sin2x = cos2x-4sin & sup2; X + 1, find TaNx

sin2x=cos2x-4(sinx)^2+1
2sinxcosx=(cosx)^2-(sinx)^2-4(sinx)^2+(sinx)^2+(cosx)^2
2sinxcosx = 2 (cosx) ^ 2-4 (SiNx) ^ 2 divided by 2 (TaNx) ^ 2
tanx=1-2(tanx)^2
TaNx = 1 / 2 or TaNx = - 1

Known: TaNx = - 2, find: (2cos ^ 2 (x / 2) - sinx-1) / radical 2Sin (π / 4 + x)
Want to know how this step (2cos ^ 2 (x / 2) - sinx-1) / root sign 2Sin (π / 4 + x) becomes (cosx SiNx) / (cosx + SiNx)!
Online, etc~

From the angle doubling formula cos & # 178; X = 2cos & # 178; (x / 2) - 1, we can get
2cos^2(x/2)-sinx-1＝cosx-sinx
From the sum angle formula, it can be concluded that
√2sin(π/4+x)=√2(sinxcosπ/4+cosxsinπ/4)＝sinx+cosx
So,
(2cos ^ 2 (x / 2) - sinx-1) / radical 2Sin (π / 4 + x)
＝(cosx-sinx)/(cosx+sinx)
＝(1-tanx)/(1+tanx)
＝(1+2)/(1-2)
＝-3

As shown in the figure, given the straight lines y = - 2 / 3x + 3 and y = 2x - 3, find the area of the triangle formed by them and y

When x = 0, Y1 = 3, y2 = - 3, the bottom is 6
When Y1 = Y2, x = 9 / 4, the height is 9 / 4
Area = 1 / 2 * 6 * 9 / 4 = 27 / 4

Curve C; y = 2x ^ 3-3x ^ 2-2x + 1. Point P (1 / 2,0), calculate the area of the figure enclosed by tangent L and C of point P

Y = 2x & sup3; - 3x & sup2; - 2x + 1y '= 6x & sup2; - 6x-2 since point P is not on the curve, let tangent point a be (a, b) and substitute point a into the curve... B = 2A & sup3; - 3A & sup2; - 2A + 1... ① y' (a) = 6A & sup2; - 6a-2 tangent passes through point P and point a, and the slope is (B-0) / (A-1 / 2) = 2B / (2a-1)

Given the quadratic function y = - 3x + C, the image passes through the point (2, - 1)
(1) Find C;
(2) How to translate the image of parabola y = - 3x & # to get the image of quadratic function?
(3) Write the vertex coordinates of the image;
(4) Is the point (- 2, - 1) on this function graph? What about the point (- 2,1)? Please explain your reason

If you take x = 2, y = - 1 into the quadratic function, you can get C = 112. If you transform the quadratic function, you can get: (Y-11) = - 3 (x-0) &# 178. According to the principle of "up minus down plus, left minus right plus", you should put the parabola y = - 3x & # 178; image up (Y-axis Square)

If the image of the line y = 3x + P intersects the X axis of the line y = - 2x + Q at the same point, then the relation between P and Q is______ ．

∵ the x-axis of the image of the line y = 3x + P and the line y = - 2x + Q intersects at the same point, ∵ when y = 0, 0 = 3x + P, x = - P3; when y = 0, 0 = - 2x + Q, x = Q2, so - P3 = Q2; when y = 0, 2p + 3Q = 0, so the answer is: 2p + 3Q = 0

Two figures of exactly the same size and shape on a plane must be symmetrical about a line?

No
Judgment of axisymmetric figure: if you fold along a straight line, the one that can coincide is the axisymmetric figure,
For example, two figures such as & amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp