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As shown in the figure, the circle C: (x-1) 2 + y2 = R2 (R & gt; 1) Let m be the intersection of the circle C and the negative half axis of the x-axis, and let m be the chord Mn of the circle C passing through M, and let its midpoint P just fall on the y-axis. (I) when r = 2, find the coordinates of the point P satisfying the condition; (II) when R ∈ (1, + ∞), find the equation of the locus g of point n; (III) the line L passing through point P (0,2) intersects with the locus g in (II) at two different points E and F, if CE · CF & gt; 0, find the value range of the slope of the line L
(1) Let n (x, y) be known, then (x-1) 2 + y2 = 4x-1 = 0, then n (1, ± 2) is obtained. So the coordinates of the midpoint P of Mn are (0, ± 1). (2): let n (x, y) be known, and let y = 0 in the circular equation, then the coordinates of m point are (1-r, 0). Let P (0, b) be obtained by kcpkmp = - 1 (or by Pythagorean theorem): r = B2 + 1 Then (x-1) 2 + y2 = r2x + 1-r = 0, eliminate R and R & gt; 1, so the trajectory equation of point n is y2 = 4x (x ≠ 0). (3) let the equation of line l be y = KX + 2, m (x1, Y1), n (X2, Y2), y = KX + 2Y2 = 4x, eliminate y to get k2x2 + (4k-4) x + 4 = 0, because line L and parabola y2 = 4x (X & gt; 0) intersect at two different points m, N, so △ = - 32K + 16 & gt; 0, so K & lt; 12, and because cm · CN & gt; 0, so (x1-1) (x2-1) + y1y2 & gt; 0, so (K2 + 1) x1x2 + (2k-1) (x1 + x2) + 5 & gt; 0, K2 + 12K & gt; 0, so K & gt; 0 or K & lt; - 12. In conclusion, 0 & lt; K & lt; 12 or K & lt; - 12 can be obtained
RELATED INFORMATIONS
- 1. If two circles intersect at points a (1,3) and B (m, - 1), and the centers of the two circles are on the straight line x + y + C = 0, then the value of M + C is () A. 0B. 2C. -3D. -1
- 2. Given two straight lines L1: MX - (2m-3) Y-1 = 0, L2: (2m + 5) x + (M + 6) Y-7 = 0, if L1 / / L2 find the value of M
- 3. If the line L1: MX + Y - (M + 1) = 0 is parallel to the line L2: x + my-2m = 0, then M=______ ．
- 4. Given the line L1: MX + Y-1 = 0 and the line L2: x + my-2m = 0, when m = 0, L1 is parallel to L2
- 5. Given the two equations L1: MX + 3Y = 0, L2 "MX + (m-2) y + = 0", when L1 is parallel to and perpendicular to L2, the range of M is calculated
- 6. It is known that two straight lines L1: MX + Y - (M + 1) = 0 and L2: x + my-2m = 0. When the value of real number m is taken, the relationship between L1 and L2 is as follows: (1) intersection; (2) coincidence; (3) perpendicularity
- 7. Let m ∈ R, X ∈ R, compare the size of x2-x + 1 and - 2m2-2mx
- 8. If x, m ∈ R, try to compare the size of X Λ 2-m + 1 and 2mx-2m Λ 2 Urgent need
- 9. M x belongs to R. compare the square of X - x + 1 with the square of - 2m - 2mx
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- 13. The area of the plane region represented by X + Y-2 ≤ 0, y ≥ x, X ≥ 0 can be found in mathematics of senior two
- 14. Given the set a = {(x, y) | x | + | y | ≤ 1}, B = {(x, y) | (Y-X) (y + x) ≤ 0}, let m = a ∩ B, then the area of the plane region corresponding to m is______ ．
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- 16. Given a > = 0, the function f (x) = (x ^ 2-2ax) e ^ X Given a > = 0, the function f (x) = (x ^ 2-2ax) e ^ X 1) When what is the value of X, f (x) gets the minimum? Prove your conclusion 2) Let f (x) be a monotone function on [- 1,1] The process needs to be clear
- 17. Let p be a point on the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) Given that point P is a point on the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), F1 and F2 are the two focuses of the ellipse, and there is a point P on the ellipse, which makes ∠ f1pf2 = 60 degree 1. Calculate the value range of ellipse eccentricity 2. Calculate the area of △ pf1f2
- 18. If XY belongs to R and has 2x + y + xy = 6, then the maximum value of 2x + y?
- 19. It is known that the line L: x-y-1 = 0 and the circle C: (x-3) square + (y-4) square = 2 are tangent to point P and pass through point P Given that the line L: x-y-1 = 0 and circle C: (x-3) square + (y-4) square = 2 are tangent to point P, the line L1 passing through point P intersects circle C at another point Q, and the length of segment PQ is 2, the L1 equation is solved
- 20. x²/m+2+y²/3-m=1 Let m = 2, the line L passing through point d (0,4) intersects curve C with two points m,. N, and O is the origin of the coordinate. If △ omn is a right triangle, the slope of L is obtained