A: Let P(x 1, y 1) be (x 1, X 1 2-2x 1+3), and I(a, b).
Ac line: y =-3x-3
Linear pa: y = (y1/(x1+1)) (x+1) = (x1-3) (x+1).
Linear PC: y = ((y1+3)/(x1)) x-3 = (x1-2) x-3.
The distance from the center to the three sides is equal, and two formulas can be written:
The distance from I to AC and PA is equal:
|-3a-3-b|/ radical symbol (1+9) =| (x1-3) (a+1)-b |/radical symbol (1+(x1)
The distance from I to AC and PC is equal:
|-3a-3-b|/ radical symbol (1+9) =| (x1-2) a-3-b)/radical symbol (1+(x 1-2) 2) ............
A and b can be solved by (1) and (2), both of which are expressed by x 1. The calculation process is complicated and is omitted here.
Finally, I got one.
2 known parabola y = ax2+x+2
(1) If the value of the algebraic expression -x2+x+2 is a positive integer, find the value of x.
(2) When a=a 1, the parabola y=ax2+x+2 intersects with the positive semi-axis of the X axis at point M(m, 0); When a=a2, the parabola y=ax2+x+2 intersects with the positive semi-axis of the X axis, and the point is N(n, 0). If point M is to the left of point N, try to compare the sizes of a 1 and a2.
Answer: (1)0 or 1,-x2+x+2 > 0, get-1
(2) Because when A = A 1, the parabola Y = AX2+X+2 intersects the positive semi-axis of the X axis at point M(m, 0).
So a 1m2+m+2 = 0 (equation 1).
Because when a = A2, the parabola y = AX2+X+2 intersects with the positive semi-axis of the X axis at point N (n, 0).
So a2n2+n+2 = 0 (Equation 2)
From Equation 2+0 of No.65438:
a 1m2+m+2=a2n2+n+2
simplify
A 1m2+m-a2n2-n=0, that is, a 1m2-a2n2+m-n=0.
Replace all m's in the equation with n!
Because point M is on the left of point N and both are on the positive semi-axis of X axis!
So m < n
Therefore, when n completely replaces m in the equation, the overall value of the equation increases.
In other words, the overall value of the equation is greater than 0.
So: a1n 2-a2n2+n-n > 0
(a 1-a2)N2 & gt； 0
And because n2>0 n2 can't be equal to 0, because it intersects the positive semi-axis of X axis and N.
So only a1-a2 >; 0
Namely a1>; Aortic second sound
3 known quadratic function Y=ax? The image of +bx+c (where a is a positive integer) passes through A (- 1, 4) and B (2, 1) and has two different intersections with the X axis, then the maximum value of b+c is ().
A: Solution: Known quadratic function Y=ax? The image of +bx+c (where a is a positive integer) passes through A (- 1, 4) and B (2, 1), so:
a-b+c=4
4a+2b+c= 1
Therefore: b=-a- 1.
c=-2a+3
Therefore: b+c=-3a+2.
Again: quadratic function Y=ax? +bx+c and x axis have two different intersections, so: △ = b * 2-4ac > 0.
Substituting b=-a- 1 c=-2a+3 gives:
(-a- 1)*2 -4a(-2a+3 )>0
Therefore: 9a * 2- 10a+ 1 > 0.
Therefore: (9a- 1) (a- 1) > 0.
Therefore: a > 1 or a < 1/9.
Also: A is a positive integer, so the minimum value of A is 2.
So b+c=-3a+2. When a takes 2, b+c has the maximum value of -4. (If the minimum value of a is 2, the maximum value of -a is -2).
4 parabola y=-(x+ 1)(x-3) is known.
The abscissas are-1 and 3, the ordinate is 3, and the center of the circumscribed circle of triangle ABC is point M (2 2,2).
Is there a point P on the parabola, which makes the straight line passing through P and M similar to the triangle BEF and triangle ABC formed by the intersection points E, F and B of AB and BC on both sides of the triangle? If it exists, find the coordinates of p; If it does not exist, explain why.
A: As long as the straight line of PM is parallel to either side, it is similar.
(1), communication in the afternoon
The analytical formula of AC is y=3x+3.
Therefore, the value of k in the analytical formula of PM is 3, and y=3x-4 is obtained by substituting m..
Combined with parabolic equation, it is found that: x1= (-1+√ 29)/2x2 = (-1-√ 29)/2.
y 1 =-5+3√29 y2 =- 1 1+3√29
P 1=(- 1+√29)/2，-5+3√29) P2=(- 1-√29)/2，- 1 1+3√29)
②BC‖PM
The analytical formula of BC is y=-x+3.
In the analytical formula of PM, the value of k is-1. If you substitute m, you get y=-x+4.
Combined with parabolic equation, it is found that x3 =-1/2+(√ 5)/2x4 =-1/2-(√ 5)/2.
y3=(9-√5)/2 y4=(7+√5)/2
P3=(- 1/2+(√5)/2，(9-√5)/2)P4= (- 1/2-(√5)/2，(7+√5)/2)
③PM‖AB(x axis)
You can get the analytical formula of PM: y=2.
Combined with parabolic equation, x5=- 1+√2 x6=- 1-√2.
y5=2 y6=2
P5=(- 1+√2，2)P6=(- 1-√2，2)
That's all for now. You can find it yourself. There is a learning network.