F is a focus of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), a and B are the two vertices of the ellipse, and the eccentricity of the ellipse is 1 / 2. Point C is on the X axis, The circle m determined by BC ⊥ BF, B, C and F is exactly tangent to the straight line L1: x + radical 3 * y + 3 = 0 1. Find the equation of ellipse 2. The straight line L2 passing through point a and circle m intersect at two points P and Q, and the vector MP * vector MQ = - 2, find the equation of straight line L2 |-c/2+b ²/ What does 2C + 3 √ (3 + 1) mean?

F is a focus of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), a and B are the two vertices of the ellipse, and the eccentricity of the ellipse is 1 / 2. Point C is on the X axis, The circle m determined by BC ⊥ BF, B, C and F is exactly tangent to the straight line L1: x + radical 3 * y + 3 = 0 1. Find the equation of ellipse 2. The straight line L2 passing through point a and circle m intersect at two points P and Q, and the vector MP * vector MQ = - 2, find the equation of straight line L2 |-c/2+b ²/ What does 2C + 3 √ (3 + 1) mean?

It can be obtained from the question that point a is on the x-axis and point B is on the y-axis. Let B (0, b) and C (x1,0), then the centrifugal force of Circle F (- C, 0) is e = 1 / 2, then a = 2C. Because BC ⊥ BF, X1 = B ²/ C X1 > 0, so the center of circle m determined by points B, C and F is (- C / 2 + b) ²/ 2c, 0) the radius is (x1 + C) / 2, and the circle m is exactly the same as the straight line

A. B is the left and right vertices of ellipse x ^ 2 / 4 + y ^ 2 / 2 = 1 A. B is the left and right vertices of the ellipse x ^ 2 / 4 + y ^ 2 / 2 = 1, passing through any t point on the straight line x = 4 to make straight lines TA and TB, and intersecting the ellipse at points m and N respectively. It is proved that point B is in the circle with Mn as the diameter

Use vector: just prove that vector MB multiplied by vector Ma is equal to 0
Point C on x = 4, C (4, Y1); a (- 2,0); B (0,2)
Then the equation of the straight line CA is listed, y = [Y1 / 6] (x + 2) is substituted into the elliptic equation, and Y1 is used to represent the x value of point M. because when solving the quadratic equation, the other solution is known to be - 2 (the x value of a in the two intersections of M and a is - 2), the x value of point m can be simply calculated by using the theorem of X1 + x2 =
In this way, m and N are represented by Y1, and the vectors MB and Ma are written. The multiplication must be 0, which is proved

Ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), A1 and A2 are the left and right vertices of ellipse C (1) Let F1 be the left focus of ellipse C, and prove that if and only if point P on C is at the left and right vertices, |pf1| obtains the minimum and maximum values (2) If the maximum value of the distance from the point on ellipse C to the focus is 3 and the minimum value is 1, the straight line L: y = KX + m intersects ellipse C at two points a and B (A and B are not left and right vertices), and aa2 is perpendicular to Ba2, verify that the straight line L passes through the vertex, and calculate the coordinates of the fixed point

1pf1 ^ 2 = (x + C) ^ 2 + y ^ 2 = (C ^ 2 / A ^ 2) (x + A ^ 2 / C) ^ 2 Pf1 = (C / a) | X - (- A ^ 2 / C) | because the shortest distance from the left vertex to the guide line - A ^ 2 / C is (a ^ 2 / C-A), the shortest distance of Pf1 is e * (a ^ 2 / C-A) = (C / a) (a ^ 2 / C-A) = a-cpf1 + PF2 = 2A, the shortest time of Pf1, the longest time of PF2 (3-1) = 2A, a = 1 x ^ 2 + y ^ 2 / b ^ 2 = 1A2 (

The limit of x ^ 2 + ax + B / X-1 is 3. Find a and B, X → 1. The steps are not very clear. Can you explain them in detail? I asked you a question last night The limit of x ^ 2 + 2x + C / x-3 is 8. Find C, X → 3. What theorem is the solution based on? When the fraction tends to a real number, the numerator and denominator are infinitesimal of the same order, and then the limit must be zero?

Because the denominator is x-3, the numerator is y = x ^ 2 + ax + B
1) When X -- > 3, the denominator is 0. To make the whole formula have a limit, it must be of type 0 / 0. Therefore:
When X -- > 3, Y -- > 0, that is, y (3) = 9 + 3A + B = 0
2) In addition, the limit of 0 / 0 type is to derive the numerator denominator respectively, and the denominator is 1 after derivation, and the numerator is y '= 2x + a after derivation. Therefore, the limit value of the whole formula is 2x + A, and this value is 8 when X -- > 3, so: 2x3 + a = 8
A and B can be solved from 1) and 2)
When the fraction approaches a limit, the numerator denominator must be infinitesimal of the same order (or infinity of the same order)

When X - > 1, if the limit of (x ^ 2 + ax + b) / sin (x ^ 2-1) is 3, find the values of a and B Specific steps, thank you Is there any other answer? I don't quite understand the 0 / 0 type. I think sin (x ^ 2-1) and (x ^ 2-1) are equivalent infinitesimal quantities, so I can replace them with (x ^ 2 + ax + b) / (x ^ 2-1) whose limit is 3 and (x ^ 2-1) whose limit is 0. Therefore, the limit of (x ^ 2 + ax + b) is also 0. I get a + B = - 1. I won't do t do t do t_ T

Since the fraction (x ^ 2 + ax + b) / (x ^ 2-1) can approach a real number and the denominator can be factorized into (x + 1) (x-1), then (x-1) must also be a factor of the molecule, because only in this way can there be a definite limit after the numerator denominator is reduced at the same time. Assuming that the numerator is reduced to (x + C), then

Find the limit a and b value x tends to 1 (x ^ 3 + ax ^ 2 + X + b) / (x ^ 2 + 1) = 3 find the AB value and find the process

a=b=2