Known circle C: (x-1) ^ 2 + (Y-2) ^ 2, through the point P (2, - 1) as the tangent of circle C, the tangent points are a and B Known circle C: (x-1) ^ 2 + (Y-2) ^ 2 = 2, through the point P (2, - 1) as the tangent of circle C, the tangent points are a and B

Known circle C: (x-1) ^ 2 + (Y-2) ^ 2, through the point P (2, - 1) as the tangent of circle C, the tangent points are a and B Known circle C: (x-1) ^ 2 + (Y-2) ^ 2 = 2, through the point P (2, - 1) as the tangent of circle C, the tangent points are a and B

Let the tangent equation be y + 1 = K (X-2), that is, kx-y - (2k + 1) = 0 tangent, that is, the distance from the center of the circle to the straight line is equal to the radius, and the center of the circle is (1,2) | K + 3 | / √ (K & sup2; + 1) = √ 2, both sides of which are square at the same time

A circle with diameter Mn just passes through the right focus of the ellipse,
Find the eccentricity of the ellipse

According to the problem, m (- C, ± 2C) is substituted into the elliptic equation, C ^ 2 / A ^ 2 + 4C ^ 2 / (a ^ 2-C ^ 2) = 1, and the solution is e = √ 2-1
The first floor answer is too complicated. For the method of calculating eccentricity of conic curve, the first choice is polar coordinates, the second choice is plane geometry, the third choice is definition, and the fourth choice is the first floor

As shown in the figure: in rectangular paper ABCD, ad = 4cm, ab = 10cm, fold as shown in the figure, so that point B and point d coincide. If the crease is EF, the de length is ()
A. 4.8B. 5C. 5.8D. 6

It is known that the function f (x) = x3-3ax2 + 2bx has a minimum value - 1 at x = 1. Try to find the value of a and B, and find the maximum value of F (x)

From the known, we can get f (1) = 1-3a + 2B = - 1, and f '(x) = 3x2-6ax + 2B, ｜ f' (1) = 3-6a + 2B = 0, ② from ①, ②, we can get a = 13, B = − 12. So the analytic expression of the function is f (x) = x3-x2-x. from this, we can get f '(x) = 3x2-2x-1. According to the properties of quadratic function, when x ＜− 13 or X ＞ 1, f' (x) ＞ 0; when − 13 ＜ x ＜ 1, f '(x) ＜ 0; Monotonically increasing on (− ∞, − 13) and (1, + ∞), monotonically decreasing on (− 13, 1) ‖ when x = − 13, f (x) has a maximum, and f (x) has a maximum of 527

Let the left and right focus of the ellipse C x ^ 2 / 9 + y ^ 2 / 4 = 1 be f 1, F 2 point p be the moving point on C, if the vector Pf1 is multiplied by the vector pf23

[parameter method] point P (3cost, 2sint) can be set by the meaning of the title, and focal point P (3cost, 2sint), and focus F1 (5,0), and focus F1 (5,0), and F2 (5,0, 5,5,0)... Vector Pf1 · vector Pf1 · vector PF2 = (3cost + √ 5,2sint) · (3cost - √ 5,2sint) = 9cos & sup2; T-5 + 4sin & sup2; t = 5cos & sup2; T-5; T-1; T-1 ＜ 0.and focus F1 (5,0, and focus (5,0), and focus F1 (5,0), and focus (5,0), and focus F1 (5,0), and focus (5,5,5,5,5, F2 (3cost) & sup2; (3 / 5; (3 / 5; (9 / 5) = = (9 / 5) = (9 / 5) = = (9 / 5); (9 / 5) = (9 / A

Given that the difference between the polynomial (- 2x's square + 3) and 2 times of a is 2x's square + 2x-7, find the polynomial a

-2x²+3-2A=2x²+2x-7
2A=-4x²-2x+10
A=-2x²-x+5

As shown in the figure, given that ball o is the inscribed ball of cube abcd-a1b1c1d1 with edge length of 1, the cross-sectional area of plane acd1 intercepting ball o is___ ．

According to the meaning of the question, plane acd1 is an equilateral triangle with side length of 2, and the tangent point of the ball and the three surfaces with point D as the common point is just the midpoint of the three sides of the triangle acd1, so the area of the cross section is the area of the inscribed circle of the equilateral triangle. Then, from the figure, the radius of the inscribed circle of △ acd1 is 22 × Tan 30 ° = 66, then the area of the cross section circle is π × 66 × 66 = π 6

1,3,7,13,21,31…… This is a group of what series, to name

n^2-n+1

PA, Pb and PC are three rays starting from point P. the angle between each two rays is 60 degrees. Then the cosine of the angle between PC and PAB is ()
A. 12B. 22C. 33D. 63

Take any point D in PC and make do ⊥ plane APB, then ⊥ DPO is the angle formed by the line PC and plane PAB. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; through point O, make OE ⊥ PA, of ⊥ Pb, because do ⊥ plane APB, then de ⊥ PA, DF ⊥ Pb. △ dep ≌ DFP, ⊥ EP =

Solve the equation 0.1 x + 0.3 x + 0.8 0.2 + 0.3 x = 8

0.5 of 0.1X + 0.3x + 0.8 of 0.2 + 0.3x = 8
0.1X/0.5+0.3X+0.2/0.8+0.3X=8
0.2X+0.3X+0.25+0.3X=8
0.2X+0.3X+0.3X=8-0.25
0.8x=7.75
x=9.6875