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4X ^ 2 + 5Y ^ 2 = 1 elliptic problem The problem is 4x ^ 2 + 5Y ^ 2 = 1 to find the length of major axis and minor axis of ellipse focus
A:
Ellipse 4x ^ 2 + 5Y ^ 2 = 1
x^2/(1/4)+y^2/(1/5)=1
a^2=1/4
b^2=1/5
So: C ^ 2 = a ^ 2-B ^ 2 = 1 / 20
The solution is as follows
a=1/2,b=√5/5,c=√5/10
So:
Long axis length 2A = 1, short axis length 2B = 2 √ 5 / 5
Focus (- √ 5 / 10,0), (√ 5 / 10,0)
It is known that the focal points of the ellipse x ^ 2 / 45 + y ^ 2 / 20 = 1 are F1 and F2 respectively. A straight line passing through the center O intersects the ellipse at two points a and B. if we want to make the area of the triangle Abf1?
It is known that the focal points of the ellipse x ^ 2 / 45 + y ^ 2 / 20 = 1 are F1 and F2 respectively. A straight line passing through the center O intersects the ellipse at two points a and B. if the area of the triangle Abf1 is 20, the equation of the straight line AB is obtained
Write out the calculation process
A:
When AB is perpendicular to the x-axis, it is easy to verify that it does not meet the problem,
Let y = KX be substituted into the coordinates of the ellipse
x^2=180/(9k^2+4),y^2=180k^2/(9k^2+4),
Then, ab = 2 √ (x ^ 2 + y ^ 2)
=2[√（k^2+1)]*180/(9k^2+4)
The distance from F1 to the line is
d=│-5*k+0*(-1)│/√（k^2+1)
S = 1 / 2 * D * ab
k=±4/3
So the linear equation is y = ± 4 / 3x
It is known that the three vertices of the triangle ABC are all on the ellipse x ^ 2 / 20 + y ^ 2 / 16 = 1, the point a is the lower end of the minor axis of the ellipse, and the center of the triangle ABC is the right focus F of the ellipse
Ellipse x ^ 2 / 20 + y ^ 2 / 16 = 1
a=2√5 b=4 c=2
Point a (0, - 4)
The center of the triangle ABC, then ABC is an equilateral triangle
The slope of af2 = 2
Then the slope of BC = - 1 / 2
Prolonging the intersection of af2 and BC at point d
AF2/F2D=2：1
Coordinates of point d (3,2)
The linear equation of BC is
y-2=(-1/2)(x-3)
x+2y-7=0
Then the slope of BC = - 1 / 2
Prolonging the intersection of af2 and BC at point d
AF2/F2D=2：1
Coordinates of point d (3,2)
The linear equation of BC is
y-2=(-1/2)(x-3)
x+2y-7=0
It is known that the image with positive scale function y = KX passes through two points a (2, - 3a) and B (3 / 2, - 9) in the fourth quadrant
Find: the coordinates of two points a and B and the positive proportion function
Because of the positive proportion function, let y = KX, K is not equal to 0
And because the image passes through the fourth quadrant, K is less than 0
That is, y = KX, K
8. It is known that the ellipse C is centered on the origin of coordinates and the axis of coordinates is the axis of symmetry, and the ellipse C is centered on the focus of parabola x ^ 2 = 16y and centered on hyperbola x ^ 2 = 16y
If the focus of Y ^ 2 / 16-x ^ 2 / 9 = 1 is the vertex, then the standard equation of ellipse C is?
X ^ 2 = 16y, focus is (0,4)
Y ^ 2 / 16-x ^ 2 / 9 = 1, focus is (0,5) and (0, - 5)
c=4,a=5
b^2=a^2-c^2=9
x^2/9+y^2/25=1
Sin29 / 6 π + cos (- 29 / 3 π) + Tan (- 25 / 4 π)
=sin(24/6+5/6)π+cos(-30/3+1/3)π+tan(-24/4-1/4)π =sin(5/6π)+cos1/3π-tan1/4π =1/2+1/2-1 =0
Time limit 5 minutes: it is known that the image with positive scale function y = KX passes through two points a (2, - 3a) and B (3 / 2, - 9) in the fourth quadrant
Find: the coordinates of two points a and B and the positive proportion function
Because of the positive proportion function, let y = KX, K is not equal to 0
And because the image passes through the fourth quadrant, K is less than 0
That is, y = KX, K
The center of ellipse and hyperbola is at the origin, the axis of symmetry is the coordinate axis, they have the same focus (25,0), and their eccentricity e can be the equation
2X ^ 2 + 4 (2e-1) x + 4E ^ 2-1 = 0 has equal real roots, find the equation of ellipse and hyperbola
If the discriminant of the equation is equal to 0, we can get E. if it is greater than 1, it is the eccentricity of hyperbola, and if it is less than 1, it is the eccentricity of ellipse. If their half focal length is 25, we can get a and B of ellipse and a and B of hyperbola, so we can get their equation
Sin (25pai / 3) + cos (- 25pai / 6) + Tan (- 29pai / 4)
sin(25π/3)=sin(8π+π/3)=sinπ/3=√3/2
cos(-25π/6)=cos(-4π-π/6)=cos(-π/6)=cosπ/6=√3/2
tan(-29π/4)=-tan29π/4=-tan(7π+π/4)=-tanπ/4=-1
And = √ 3-1
If the image of positive scale function y = KX + B (K ≠ 0) passes through the fourth quadrant, and passes through two points (2, - 3a) and (- A, 6)
Find the value of the independent variable x when the function value y = - 3
If B is a positive proportional function, then B = 0,
We can get k = 3 or K = - 3 by taking the point into the function,
And the image goes through the fourth quadrant,
Then k = - 3,
So when y = - 3, x = 1
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