The left and right fixed points of ellipse C1: x ^ 2 / A ^ 2 + y ^ 2 / 4 = 1 (a is greater than 2) are a and B respectively, and point P is the image of hyperbola C2: x ^ 2 / A ^ 2-y ^ 2 / 4 = 1 in the first quadrant If the area of triangle ACD and triangle PCD is equal, (1) calculate the coordinates of point p; (2) whether the line CD can pass the right focus of ellipse C1, if it can, calculate the value of A. if not, please explain the reason

2. The left quasilinear l of hyperbola C1; X ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1, F1F2 is the left and right focus respectively, F1F2 = 2 * radical (41) = 2 * radical 5, if pf1f2 is a right triangle, so Po = f1o = f2o so

Line AB is a diameter of circle C1: x2 + Y2 + 2x-6y = 0, and the focus of hyperbola C2 with eccentricity 5 is a and B. if P is a common point of circle C1 and hyperbola C2, then | PA | + | Pb | = ()
A. 22B. 42C. 43D. 62

The radius r = 124 + 36 = 10 of c1:x2 + Y2 + 2x-6y = 0 radius r = 124 + 36 = 10, line AB is a diameter of c1:x2 + Y2 + 2x-6y = 0, and hyperbola C2 with centrifugal rate of 5 focuses on a and B with a, B as the focus of a, B as the focus, the focal length of hyperbola C2 is 2C = | ab | P is a common point of circle C1 and hyperbola C2, and a common point of circle C1 and hyperbola C2, and a common point of circle C1 and hyperbola is a common point of circle C1 and hyperbola, and a common point of circle C1 and hyperbola and hyperbola is a common point of circle C1 and hyperbola and hyperbola, such as Pb |||||||||||||||||124; Pb | = 4a2, ∵ C = 10, e = CA = 5, | a = 2, | 2 | PA | Pb | = 32, | Therefore, D

It is known that the line y = 2x-3 intersects with the x-axis at point a, and intersects with the y-axis at point B, while the line y = KX + B passes through point B and point C, where point C and point a are symmetric about the y-axis, and the values of K and B are obtained

The line y = 2x-3 intersects the X axis at point a, let y = 0, x = 3 / 2, so point a is (3 / 2,0) the line y = 2x-3 intersects the Y axis at point B, let x = 0, y = - 3, so point B is (0, - 3) because point C and point a are symmetric about the Y axis, so point C is (- 3 / 2,0). Substitute two points of B and C into the line y = KX + B respectively - 3 = k * 0 + B0 = k * (- 3 / 2) + B solution

If the line y = x is the tangent of the curve y = x3-3x2 + ax, then a=______ .
The third power of X is three times the square of X
If you think about it again, the solution of X is 0 and 3 / 2, and a is 1 and 13 / 4

Let the tangent point be (x0, Y0)
If the tangent point is on the tangent y = x, then: Y0 = x0
If the tangent point is on the curve y = x ^ 3-3 * x ^ 2 + A * x, then: Y0 = x0 ^ 3-3 * x0 ^ 2 + A * x0
And y '= 3 * x ^ 2-6 * x,
Then y '| (x = x0) = 3 * x0 ^ 2-6 * x0 + a = 1, then
x0(x0^2-3*x0+a-1)=0;
x0^2-2x0+(a-1)/3=0;
be
x0=0;
x0^2-2x0+(a-1)/3=0;
The solution is as follows
x0=0;
a=1
or
x0^2-3*x0+a-1=0;
x0^2-2x0+(a-1)/3=0;
The solution is as follows
A = 3 * x0 / 2 + 1 generation
x0^2-3*x0+3/2*x0=0;
X0 = 0 or x0 = 3 / 2;
Accordingly, a = 1 or a = 13 / 4
Review; a = 1 or a = 13 / 4

If the function f (x) = x3-3x2 + 1 is known, then the tangent equation with the smallest slope is______ ．

F '(x) = 3x2-6x = 3 (x-1) 2-3, when x = 1, the minimum value of F' (x) is - 3, that is, the minimum value of the slope is - 3, and f (1) = 1-3 + 1 = - 1, then the tangent point is (1, - 1), and the tangent equation of the minimum slope is Y - (- 1) = - 3 (x-1), that is, 3x + Y-2 = 0, so the answer is: 3x + Y-2 = 0

Among the tangents of the curve y = X3 + 3x2 + 6x-10, the equation with the smallest slope is

Y '= 3 (x ^ 2 + 2x + 2)
When x = - 1, y 'has a minimum value of 3
That is to say, at (- 1, - 14), the tangent slope is the smallest, and the slope is 3
y-(-14)=3(x-(-1))
After sorting out, we get: 3x-y-11 = 0, which is the result

Given that the line y = KX and the curve y = 3x ^ 3 + 2x ^ 2 are tangent to the point (x0, Y0), then the tangent point coordinates are

At the tangent point, the slope of the curve is equal to that of the straight line, and the derivative of the curve is y '= 9x2 + 4x = K. The equation can be solved by the formula method, and the corresponding (x0, Y0) can be obtained,

The area of the figure enclosed by the line L perpendicular to the line 2x-6y + 1 = 0 and tangent to the curve f (x) = X3 + 3x2-1 and the curve f (x) and Y axis is______ ．

Let the equation of the straight line l be y = - 3x + m and the tangent point be (n, N3 + 3n2-1), then 3N2 + 6N = - 3, n = - 1 can be obtained from the meaning of the title, so the tangent point is (- 1, 1), which is substituted into the tangent equation If y = - 3x + m, M = - 2 can be obtained. The equation of line L is y = - 3x-2. The area of the figure enclosed by line L, curve f (x) and Y axis is ∫ 0 − 1 [(X3 + 3x2 − 1) − (− 3x − 2)] DX = ∫ 0 − 1 (x + 1) 3DX = 14 (x + 1) 4 | 0 − 1 = 14, so the answer is: 14

The tangent equation of the curve y = x3-3x2 + 1 parallel to the straight line 3x + Y-10 = 0 is______ ．

Let the tangent parallel to the straight line 3x + Y-10 = 0 and tangent to the curve y = x3-3x2 + 1 and the tangent point of the curve be (x0, X03 − 3x02 + 1). From y = x3-3x2 + 1, y ′ = 3x2-6x, then y ′| x = x0 = 3x02 − 6x0. So 3x02 − 6x0 = − 3, that is, X02 − 2x0 + 1 = 0, so x0 = 1. Then X03 − 3x02 + 1 = 13 − 3 × 12 + 1 = − 1. So the tangent point is (1, - 1). So the tangent equation is Y - (- 1) = - 3 × (x-1) ）The answer is 3x + Y-2 = 0

If there are two different intersections between the line y = KX + 2 and the circle (X-2) (X-2) + (Y-3) (Y-3) = 1, then the value range of K is?

If y = KX + 2 is substituted into the following equation, then (X-2) (X-2) + (kx-1) (kx-1) = 1, then (k * k + 1) x ^ 2 - (2k + 4) x + 4 = 0, then the discriminant △ = (2k + 4) ^ 2-4 * 4 (k ^ 2 + 1) = - 12K ^ 2 + 16K is greater than 0, so the range of K is 0

Line AB is a diameter of circle C1: x2 + Y2 + 2x-6y = 0, and the focus of hyperbola C2 with eccentricity 5 is a and B. if P is a common point of circle C1 and hyperbola C2, then | PA | + | Pb | = () A. 22B. 42C. 43D. 62