(1) Let AB be the chord passing through the center of the ellipse x * 2 / A * 2 + y * 2 / b * 2 = 1 (a > b > 0), and the left focus of the ellipse is F1 (- C, 0), then Δ F (1) Let AB be the chord passing through the center of the ellipse x * 2 / A * 2 + y * 2 / b * 2 = 1 (a > b > 0), and the left focus of the ellipse is F1 (- C, 0), then the area of Δ f1ab is the largest——————

BC
It is proved that: when the slope of a straight line passing through the origin does not exist, the triangle area is BC, when the slope exists, it is set as K, the coordinates of two intersections are set as a (x1, Y1) B (X2, Y2), and the linear equation y = KX is brought into the elliptic equation to get (A & sup2; K & sup2; + B & sup2;) x & sup2; - A & sup2; B & sup2; = 0,
x1+x2=0,x1x2=-a²b²／（a²k²＋b²）,y1+y2=0 y1y2=-k²a²b²／（a²k²＋b²）,
Triangle area = 0.5c | y1-y2 | = 0.5 √ (4K & sup2; a & sup2; B & sup2; / (A & sup2; K & sup2; + B & sup2;)) = √ [K & sup2; a & sup2; B & sup2; / (A & sup2; K & sup2; + B & sup2;)]
=B √ [1 / (1 + B & sup2; / A & sup2; K & sup2;)], the larger the K & sup2; is, the larger the area is. When K does not exist, the largest area is BC
It is known that F 1 and F 2 are two focal points of ellipse x 2 + y 22 = 1, and ab is a moving chord passing through the focal point F 1
∵ F1 and F2 are the two focal points of the ellipse x2 + Y22 = 1, ∵ F1 (0, - 1), a = 2, B = C = 1, ∵ AB is a moving chord passing through the focal point F1, ∵ rotate the line AB around the F1 point. According to the geometric properties of the ellipse, it is obtained that when AB is perpendicular to the major axis of the ellipse, ∵ abf2 takes the maximum area, and ∵ abf2 takes the maximum area s
The focus of the parabola ^ 2-2 / y is at the origin of the parabola
The vertex of the parabola is at the origin, and its quasilinear passes through the left focus of the hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1, and is perpendicular to the X axis. The parabola and the curve intersect at (3 / 2, root sign 6). The equation of the parabola and the hyperbola is obtained
Let the parabolic equation be y2 = 4C and X,
∵ parabola passing point (3 / 2, √ 6), ∵ 6 = 4C &; 3 / 2
Therefore, the parabolic equation is y2 = 4x
The hyperbola x2 / A2 - Y2 / B2 = 1 passes through the point (3 / 2, √ 6),
9 / 4a2 - 6 / B2 = 1, A2 + B2 = C2 = 1, 9 / 4a2 - 6 / 1-a2 = 1
A 2 = 1 / 4 or a 2 = 9 (rounding)
∴b2= 3/4,
4x2- 4y2/3=1．
By simplifying cos ^ 2 (quarter - α) - Sin ^ 2 (quarter - α), we get
cos^2（π/4-α）-sin^2（π/4-α）
={cos（π/4-α）+sin（π/4-α）} {cos（π/4-α）-sin（π/4-α) }
=Radical 2 {cos π / 4 cos (π / 4 - α) + sin π / 4 sin (π / 4 - α)} radical 2 {cos π / 4 cos (π / 4 - α) - sin π / 4 sin (π / 4 - α)}
=2 cos[π/4-(π/4-α）] {cos[π/4+(π/4-α)]
=2 cosα cos(π/2-α)
=2 cosα sinα
=sin2α
If the image passes through the first quadrant and the third quadrant, what is k equal to,
K is less than 0
Because in the positive scale function, when k is greater than 0, the image passes through one or three quadrants; when k is less than 0, the image passes through two or four quadrants,
The positive proportion function y = - KX passes through one or three quadrants, so - K should be greater than 0, so K should be less than 0
K is less than or equal to 0
Because it passes through one or three quadrants - K is greater than 0, and K is not equal to 0, so it is less than 0
Find the hyperbolic equation and eccentricity with the midpoint at the origin, the axis of symmetry as the coordinate axis, a focus (- 4,0), and the daily asymptote 3x-2y = 0
Hyperbolic equation
(x²/a²)-(y²/b²)=1
We can get it from the problem
a²+b²=c²
C=4
b∶a=3∶2
e=c/a
The solution is as follows
a=8/√13,
b=12/√13
C=4
e=(√13)/2
The hyperbolic equation
(13x²/64)-(13y²/144)=1
Eccentricity e = (√ 13) / 2
Simplify sin (- A-7). Cos (A-3 / 2)=
-sin²α
Solution sin (- A-7). Cos (A-3 / 2)
=-Sin (a + 7). Cos (3 / 2-A)
=-Sin (a + V). Cos (3 V / 2-A)
=sina×(-sina)
=-sin^2a
The images with known positive scale function pass through quadrants 1 and 3, and pass through two points (2, 3a) and (a, 6)
It is known that the image of a positive scale function passes through quadrants 1 and 3 and passes through two points (2,3a) and (a, 6). The analytic expression of the function is obtained. When the value of the function is 6, the value of the independent variable x is as shown in the figure. It is known that a (3,0), the line passing through point a and perpendicular to the x-axis intersects with the line y = x at point B. in the plane rectangular coordinate system, the image of the function in (1) intersects with the line passing through point a and perpendicular to the x-axis at point C, Finding the area of △ OBC
When y = KX + B is brought into three coordinate points to solve the equation, k = 3, B = 0, a = 2, so the analytic expression of the function is y = 3x, when the function value is 6, the independent variable x = 2
The second question is unclear
It is known that a focus of the hyperbola whose center is at the origin is (- 4.. 0). An asymptotic equation is 3x-2y = 0, which is used to solve the hyperbolic equation
Because the focus of hyperbola is (- 4,0)
Let the hyperbolic equation be
(x^2/a^2)-[y^2/(16-a^2)]=1
The asymptote equation is y = (3 / 2) X
So B ^ 2 / A ^ 2 = 9 / 4
That is, 16 / (a ^ 2) = 13 / 4
a^2=64/13
The hyperbolic equation is (13X ^ 2 / 64) - (144y ^ 2 / 64) = 1
sin29π/6+cos(-29π/3)+tan(-25/4)
sin29π/6+cos(-29π/3)+tan(-25π/4) =sin(24π/6+5π/6)+cos(-30π/3+π/3)+tan(-24π/4-π/4) =sin5π/6+cosπ/3-tanπ/4

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