Given that the vertices B and C of △ ABC are on the ellipse x23 + y2 = 1, vertex A is a focus of the ellipse, and the other focus of the ellipse is on the edge of BC, then the perimeter of △ ABC is () A. 23B. 6C. 43D. 12

Given that the vertices B and C of △ ABC are on the ellipse x23 + y2 = 1, vertex A is a focus of the ellipse, and the other focus of the ellipse is on the edge of BC, then the perimeter of △ ABC is () A. 23B. 6C. 43D. 12

From the definition of ellipse: the sum of the distances from a point on the ellipse to two focal points is equal to the length of the major axis 2a, we can get that the perimeter of △ ABC is 4A = 43, so we choose C
It is known that the vertices B and C of triangle ABC are on the ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1, vertex A is a focus of the ellipse, and the other focus of the ellipse is on the edge of BC,
What is the perimeter of the triangle ABC?
Let the left focus of the ellipse be F1, and the right focus be F2. Then the perimeter of the triangle ABC is equal to the perimeter of the triangle bf1f2 plus the perimeter of the triangle cf1f2. According to the elliptic equation, we know that 2A = 4  the perimeter of the triangle ABC = 4 + 4 = 8
It is known that the vertices B and C of the triangle ABC are on the ellipse x ^ 2 / 3 + y ^ 2 = 1. The vertex A is a pair of the ellipse, which should be twice the length of the major axis, that is, 4A a = √ 3, so the perimeter is 4 √ 3
Let the intersection point of BC be D, defined by ellipse, AB + BD = AC + CD = 4,
So the perimeter of triangle ABC is ab + BD + AC + CD = 8
Given that x is the third quadrant angle and sin ^ 4x + cos ^ 4x = 5 / 9, sin2x is equal to
Let a = sin2xsin ^ 4x + cos ^ 4x = (sin ^ 2x + cos ^ 2x) ^ 2-2sin ^ 2xcos ^ 2x = 1 - (2sinxcosx) ^ 2 / 2 = 1 - (sin2x) ^ 2 / 2 = 5 / 9, then sin2x = radical 2 * (2 / 3) or - radical 2 * (2 / 3) because x is in the third quadrant, then 2x is in the first and second quadrants, sin2x > 0, so sin2x = radical 2 * (2 / 3)
2 / 3 times the root 2
It's very simple
sin^4x+cos^4x = (sin^2x+cos^2x)^2-2sin^2x*cos^2x
=1-((2sinx*cosx)^2)/2=1-(sin^2 2X)/2=5/9
So sin2x = 8 / 9 open root
SIN^4X+COS^4X=(sin^2x+cos^2x)^2-2sin^2x*cos^2x=1-0.5sin^2(2x)
=5/9
sin^2(2x)=8/9
PI
The image of positive scale function y = KX (k is constant, K is not equal to 0) is processed by?
(0,0)(1,k)
A line passing through the origin
The image of y = KX is a straight line passing through (0,0) and (1, K).
Why is the distance from the focus of hyperbola to the origin multiplied by the sine of asymptote and X-axis equal to the distance from the vertex to the origin?
The distance from the focus to the origin is equal to C,
The tangent of the angle between the asymptote and the x-axis is B / A, so the cosine is a / C
The product is equal to a, that is, the distance from the vertex to the origin
There is a mistake in your question,
In addition, a right triangle can be constructed, and a hyperbolic tangent can be made through the vertex to intersect any asymptote
It can be proved that the three sides of this right triangle are a, B and C respectively
Sin ^ 4x + cos ^ 4x = 5 / 9 for sin 2x
Answer process
sin^4x+cos^4x=5/9
(sin^2x+cos^2x)^2-2sin^2xcos^2x=5/9
1-sin^2(2x)/2=5/9
sin^2(2x)=8/9
Sin2x = plus or minus 2 root sign 2 / 3
Let a = sin2x
SIN^4X+COS^4X=(sin^2x+cos^2x)^2-2sin^2xcos^2x
=1-(2sinxcosx)^2/2=1-(sin2x)^2/2=5/9
Then sin2x = radical 2 * (2 / 3) or - radical 2 * (2 / 3)
Because x is in the third quadrant, then 2x is in quadrant 1 and 2, sin2x > 0
So sin2x = root 2 * (2 / 3)
Why can't K be equal to 0 in the positive scale function "y = KX (k is a constant, K ≠ 0)"?
Because the definition of positive scale function is:
F (x) is defined as a positive proportional function, if for any X and y, f (x) satisfies
f(x)/f(y)=x/y.
If k = 0 in F (x) = KX,
Then f (x) / F (y) is meaningless
Therefore, it is necessary to declare K ≠ 0
If k = 0, then it is meaningless to take any value of X in the positive proportional function, the result is always equal to 0, and the function becomes y = 0. Is it meaningful for X to be a variable
Because it's proportional
If k = 0, it becomes a straight line y = 0
If there is a point m on the equiaxed hyperbola and the distance from the origin is 2, what is the distance from m to the two focal points
Because a = B
So let the equation be x ^ 2-y ^ 2 = a ^ 2
Think x ^ 2 + y ^ 2 = 4
So 2x ^ 2 = 4 + A ^ 2
y^2=4-2-0.5a^2=2-0.5a^2
Because the point is on the hyperbola
2+0.5a^2+2-0.5a^2=a^2
a^2=4
d1-d2=4
1) Find ^ cosin = 4x / 4sin
The more detailed the process, the better
1. Sin ^ 4x + cos ^ 4x = (sin ^ 2x + cos ^ 2x) ^ 2 - 2Sin ^ 2xcos ^ 2x = 5 / 8 ∵ sin ^ 2x + cos ^ 2x = 1 ∵ 1-2sin ^ 2xcos ^ 2x = 5 / 82sin ^ 2xcos ^ 2x = 3 / 8, then: (sin ^ 2x) ^ 2 = 3 / 4cos4x = 1-2 (sin 2x) ^ 2 = 1 - 3 / 2 = - 1 / 22. ∵ sin ^ 4x cos ^ 4x = (sin ^ 2x + cos ^ 2x) (sin ^ 2x -
①sin²x+cos²x=1
Square rule on both sides
（sin^4 x+cos^4 x）+2（sin²xcos²x）=1
Then Sin & sup2; xcos & sup2; X = 3 / 16
cos4x=cos²2x-sin²2x
=(COS & sup2; x-sin & sup2; x) & sup2; - 4sin & sup2; xcos... Expansion
①sin²x+cos²x=1
Square rule on both sides
（sin^4 x+cos^4 x）+2（sin²xcos²x）=1
Then Sin & sup2; xcos & sup2; X = 3 / 16
cos4x=cos²2x-sin²2x
=（cos²x-sin²x）²-4sin²xcos²x
=sin^4x+cos^4x - 6sin²xcos²x
=5/8-9/8 = -1/2
② sin^4x-cos^4x=
（sin²x+cos²x）（sin²x-cos²x）=sin²x-cos²x=-4/5
Then cos2x = cos & sup2; x-sin & sup2; X = 4 / 5
So sin2x = 3 / 5 or - 3 / 5? Fold up
Why is the image of positive scale function y = KX (k is constant, K is not equal to 0) a straight line passing through two points (0,0), (1, K)
First, two points determine a straight line;
Secondly, (0,0), (1, K) are on the linear equation y = KX;
So the image with positive scale function y = KX (k is constant, K is not equal to 0) is a straight line passing through two points (0,0), (1, K)
First, two points determine a straight line;
Secondly, (0,0), (1, K) are on the linear equation y = KX;
So the image with positive scale function y = KX (k is constant, K is not equal to 0) is a straight line passing through two points (0,0), (1, K).