If we know that the vertex of the triangle is parallel to the origin of the triangle, The same as above, find the area of triangle ABC

If we know that the vertex of the triangle is parallel to the origin of the triangle, The same as above, find the area of triangle ABC

If we use the formula AB = 1, we can find out the slope of the point where the ellipse lies, Don't you mind?
It is known that the left and right focal points of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a greater than 0, b greater than 0) are F1F2, the eccentricity e = (radical 2) / 2, and the equation of the right guide line is x = 2
The standard equation of ellipse
The eccentricity e = (radical 2) / 2, and the equation of right collimator is x = 2
∴ c/a=√2/2, a²/c=2
∴ a=√2c, a²/c=2
∴ 2c²/c=2
∴ c=1
∴ a=√2
∴ b²=a²-c²=1
The standard equation of ellipse is X & # 178 / 2 + Y & # 178; = 1
The center of the ellipse is at the origin, a vertex is (0,2), eccentricity e = root 3 / 2, find the elliptic equation
The center of the ellipse is at the origin, a vertex is (0,2), eccentricity e = (√ 3) / 2, find the elliptic equation
① When the focus is on the y-axis, it has a = 2, e = C / a = C / 2 = (√ 3) / 2, so C = √ 3, B & # 178; = A & # 178; - C & # 178; = 4-3 = 1
So the elliptic equation is: Y & # 178 / 4 + X & # 178; = 1
② When the focus is on the x-axis, according to the meaning of the title, there is b = 2, from a & # 178; - B & # 178; = C & # 178;, 1 - (B / a) &# 178; = (C / a) &# 178; = E & # 178; = 3 / 4,
So (B / a) &# 178; = 1-3 / 4 = 1 / 4, substituting B = 2 into a & # 178; = 4 / (1 / 4) = 16,
So the elliptic equation is: X & # 178 / 16 + Y & # 178 / 4 = 1
Analysis:
Because the center of the ellipse is at the origin, a vertex is (0, 2), that is, the focus is on the Y axis, and a = 2.
And because eccentricity e = root 3 / 2, so C = √ 3, so B = 1
So the elliptic equation is: y ^ 2 / 4 + x ^ 2 = 1
The vertex is on the Y axis, so
If the focus is on the y-axis, then 2 = a
C / a = radical 3 / 2
C = root 3
b^2=a^2-c^2=4-3=1
y^2/a^2+x^2/b^2=1
y^2/4+x^2/1=1
Draw the image and range of y equal to the absolute value of x plus 1 minus the absolute value of x minus 1
. . { 2 x>1
y=|x＋1|－|x－1| = { 2x －1≤x≤1
{ －2 x
The discussion of X classification can be divided into the following three cases: X1 discussion. There are three kinds of corresponding Y: y = - 2 (x)
Find the real axis length, imaginary axis length, focal length, vertex coordinate, eccentricity and asymptote equation of hyperbola 16 / x square - 9 / y square = 1
χ & # 178 / 16 - γ & # 178 / 9 A = 4 B = 3 real axis: 2A = 8 imaginary axis: 2B = 6 C = = √ A & # 178; + B & # 178; = √ 16 + 9 = 5 focal length: 2C = 10 focal point: F1 (- 5,0) F2 (5,0) vertex coordinate: A1 (- 4,0) A2 (4,0) eccentricity: e = C / a = 5 / 4 asymptotic equation: y = ± B / a χ = ± 3 / 4 χ
Given that sin (π - X of 4) = 3 of 5, then sin 2x,
∵sin（π/4-x）=3/5
∴sinπ/4cosx-cosπ/4sinx=3/5
That is √ 2 / 2 * cosx - √ 2 / 2sinx = 3 / 5
∴cosx-sinx=3√2/5
The square of both sides is: cos & # 178; X + Sin & # 178; x-2sinxcosx = 18 / 25
That is 1-sin2x = 18 / 25
∴sin2x=1-18/25=7/25
How to draw function image with absolute value
For example, what are the drawing methods of y = | x | + 1 and y = | x + 1?
Are all graphs drawn without absolute values, and then fold the components below the x-axis along the x-axis?
For example, y = | x | + 1
First draw y = | X|
Draw y = | x | draw the image of y = x, and then fold up the part of y = x below the x-axis along the x-axis, with the upper part unchanged to get y = | X|
Y = | x | + 1 is obtained by translating the y = | x | - image up one unit
y=|x+1|
Draw the image of y = x + 1, then fold the part of y = x + below the x-axis up along the x-axis, and keep the upper part unchanged to get y = | x + 1|
Are all graphs drawn without absolute values, and then fold the components below the x-axis along the x-axis?
This is not absolute
For example, if you draw y = LG | x |, you can draw y = lgx and fold it along the y-axis
For the case of absolute value, we should first remove the sign of absolute value (with the help of the absolute value of non negative number as itself, and then discuss the situation), generally we can get the piecewise function, and then discuss the piecewise function.
Find the real axis length, imaginary axis length, focal length, vertex, coordinate, focus coordinate, eccentricity, and asymptote equation of hyperbola y-4x square = 1
y²－4x²=1
y²－x²/(1/4)=1
Then a & # 178; = 1, B & # 178; = 1 / 4, then C & # 178; = A & # 178; + B & # 178; = 5 / 4, then C = √ 5 / 2
The real axis is 2A = 2, the imaginary axis is 2B = 1, the vertex of the real axis is (0,1), (0, - 1), the vertex of the imaginary axis is (1 / 2,0), (- 1 / 2,0); the focal length is 2C = √ 5; the focus is (0, √ 5 / 2), (0, - 5 / 2); the eccentricity is e = C / a = √ 5 / 2; the asymptote is y = 2x or y = - 2x
It's class time now. How to ask questions here
If sin (PAI / 4-x) = 3 / 5, what is the value of sin2x
Can you write down the process in detail?
sin(∏/4-x)=sin∏/4cosx-cos∏/4sinx
=(radical 2) / 2 (cosx SiNx) = 3 / 5
Square of both sides: 1 / 2 (1-2cosxsinx) = 9 / 25
Namely: 1-sin2x = 18 / 25
sin2x=7/25
Sin (PI / 4-x) = 3 / 5, cos (PI / 2-2x) = 7 / 25, then cos (2x pi / 2) = 7 / 25, from odd to even constant, sign quadrant sin2x = 7 / 25
Quadratic function adds absolute value to a term, so how to draw function image
Seek the picture! Or the text also may thank!
Draw in sections, divide x > = 0 and X