It is known that vertex BC of triangle ABC is on ellipse x ^ 2 / 3 + y ^ 2 = 1, vertex A is one focus of ellipse, and the other focus of ellipse is on edge BC to find triangle ab Girth of

It is known that vertex BC of triangle ABC is on ellipse x ^ 2 / 3 + y ^ 2 = 1, vertex A is one focus of ellipse, and the other focus of ellipse is on edge BC to find triangle ab Girth of

Hello ~ this is a basic problem ~ to study the definition of ellipse: to the set of points whose focus is fixed value 2A (2a > | F1F2 |)
The perimeter of triangle ABC can be divided into two parts: (let focus a, f) one is | ab | + | BF | = 2A, the other is | AC | + | CF | = 2A
So the perimeter of the triangle = 2A + 2A = 4A = 4, root 3
Thank you~
Suppose that the center of the ellipse is at the origin of the coordinate, the focus is on the x-axis, a vertex (2,0), and the eccentricity is 3 / 2 of the root. If the left focus of the ellipse is F1 and the right focus of the ellipse is F2, and a straight line passing through F1 with a slope of 1 intersects the ellipse at B, calculate the area of △ abf2
First of all, it is easy to get the elliptic equation a = 2, B = 1, C = root 3; X ^ 2 / 4 + y ^ 2 = 1
F1 (- radical 3,0), straight line; y = x + radical 3, substituting into elliptic equation (eliminating x)
That is, 5Y ^ 2-2 radical 3y-1 = 0
The solution is Y1, Y2
The area of △ abf2 = 1 / 2af2 times the absolute value of Y
Where af2 = 2-radical 3, the absolute value of Y is obtained by solving the equation
In the plane rectangular coordinate system xoy, if △ ABC vertex a (- 4,0) and C (4,0) are known, and vertex B is on the ellipse X225 + Y29 = 1, then Sina + sincsinb = ()
A. 34B. 23C. 45D. 54
In the ellipse X225 + Y29 = 1, a = 5, B = 3, C = 4, so a (- 4,0) and C (4,0) are the two focal points of the ellipse, ∧ AB + BC = 2A = 10, AC = 8, according to the sine theorem, Asina = bsinb = csinc = 2R, ∧ Sina + sincsinb = a + CB = AB & nbsp; + & nbsp; BCAC = 108 = 54, so D
Find the equation of equiaxed hyperbola which passes through point a (3, - 1) and whose symmetry axis is on the coordinate axis
When the focus is on the x-axis, let the standard equation of hyperbola be x2a2 − y2a2 = 1, substituting a (3, - 1) into the equation to get 9A2 − 1A2 = 1, A2 = 8, and the standard equation of hyperbola be x28 − Y28 = 1
Given sin (π / 4-x) = 5 / 13, 0 < x < π / 4, find the value of sin 2x / cos (π / 4-x)
Need to use double angle formula.. not familiar with TAT
The beauty of Mathematics
sin(π/4 - x) = 5/13
0 < x < π/4
0 < π/4 - x < π/4
cos(π/4 - x) > 0
∴cos(π/4 -x) = 12/13
sin(2x)
= cos(π/2 - 2x)
= cos 2(π/4 -x)
= 2cos²(π/4 -x) - 1
sin(2x) / cos(π/4 - x)
= 2cos(π/4 -x) －1/cos(π/4 - x)
= 2×12/13 - 13/12
= 119/156
If the value of the independent variable of the positive proportion function y = KX increases by 1, the value of the function will decrease by 2 correspondingly. Find the analytic formula of the function (find the whole process 0)
y=-2x
When x = 1, y = K
When the value of x increases by 1, that is x = 2, y = 2K,
Because the value of function 2 decreases accordingly,
That's 2K = K-2, so k = - 2
So the analytic expression is y = - 2x
-2 = y2-y1 = kx1-kx2 = K (x1-x2) = k * 1 = k, so y = - 2x
P
Y = - 2x, oh, yes? Follow up question: This is the eighth grade math problem
The center is at the origin, the axis of symmetry is on the coordinate axis, passing through the points m (3,15 / 4), n (16 / 3,5), the hyperbolic equation is solved
The center is at the origin, which means that the asymptote is symmetric about the origin, C = 0. Two points determine the hyperbolic equation, which can be solved
It is proved that 2Sin (π + x) cos (π - x) = sin2x
prove:
The induction formula and double angle formula can be used
Left = 2Sin (π + x) cos (π - x)
=2*(-sinx)*(-cosx)
=2sinxcosx
=sin2x
=Right
∴ 2sin(π+x)cos(π-x)=sin2x
The image with positive scale function y = KX (k is constant, K is not equal to 0) is processed_____ ,______ A straight line between two points
(0,0)(1,k)
(0,0),(1,k)
If there is a point on an equiaxed hyperbola where the distance between the origin of the M derivative coordinate is 2, what is the product of the distances between the focus of the M derivative
Equiaxed hyperbola means a = B. the standard equation of hyperbola is: x ^ / A ^ - y ^ / A ^ = 1C = √ (a ^ + A ^) = √ 2A. The two focuses of hyperbola are: F1 (- √ 2a, 0), F2 (√ 2a, 0). Let the coordinates of m point be: (m, n) from the meaning "the distance from m to the origin is 2", we can get: m ^ + n ^ = 2 ^ = 4