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In the plane rectangular coordinate system xoy, the square of the straight line L and the parabola y = 2x intersects at two points A.B
In the plane rectangular coordinate system xoy, the straight line L and the parabola y ^ 2 = 2x intersect at two points a and B. (1) verification: "if the straight line L passes through the point t (3,0), then OA.OB=3 It is true proposition (2). Write the inverse proposition of the proposition in (1) (the intersection of the straight line L and the parabola y ^ 2 = 2x at two points a and B is the major premise), and judge whether it is a true proposition or a false proposition, Write the proof process; if it is a false proposition, give a counter example
Is this the problem?
When K does not exist, AB: x = 3
Suppose a is above and B is below, let a (x1, Y1) B (X2, Y2)
Then X1 = x2 = 3, Y1 = 6 ^ (1 / 2), y2 = - 6 ^ (1 / 2)
therefore OA.OB=X1 .X2+Y1.Y2=9-6=3
When k exists, AB: y = K (x-3)
When y ^ 2 = 2x, the
(K / 2). Y ^ 2-y-3k = 0, K is not equal to 0
Then, from the Veda theorem Y1. Y2 = - 6
be OA.OB=X1 .X2+Y1.Y2=(Y1Y2)^2/4+Y1Y2=9-6=3
false
Counterexample: y = x / 2 + 1 / 2 and Y ^ 2 = 2x intersect a (x1, Y1), B (X2, Y2)
Then a (3 + 2.2 ^ (1 / 2), 2 + 2 ^ (1 / 2)) B (3-2.2 ^ (1 / 2), 2-2 ^ (1 / 2))
be OA.OB= (9-8)+(4-2)=3
In this case, the intersection with the X axis at (- 1,0) is not (3,0)
So it's false
As shown in the figure, it is known that in the plane rectangular coordinate system xoy, the parabola y = AX2 + BX + C (a > 0) intersects the X axis at two points a (- 1,0) and B (3,0) The symmetry axis L intersects with X axis at point C, the vertex is point D, and the tangent of ∠ ADC is 1 2． (1) Find the coordinates of vertex D; (2) Find the expression of parabola; (3) Point F is a point on a parabola, and it is located in the first quadrant. Connect AF. if ∠ fac = ∠ ADC, find the coordinates of F point
(1) ∵ the parabola intersects the X axis at two points a (- 1,0) and B (3,0),
﹣ the symmetry axis straight line L = - 1 + 3
2=1，
∵ the symmetry axis L intersects the X axis at point C,
∴AC=2，
∵∠ACD=90°，tan∠ADC=1
2，
∴CD=4，
∵a＞0，
∴D（1，-4）；
(2) Let y = a (X-H) 2 + K, and (1) we know that h = 1, k = - 4,
∴y=a（x-1）2-4，
Put x = - 1, y = 0 into the above formula,
A = 1,
Therefore, the parabola is expressed as y = x2-2x-3;
(3) ⊥ h is the vertical point,
Let f (x, x2-2x-3),
∵∠FAC=∠ADC，
∴tan∠FAC=tan∠ADC，
∵tan∠ADC=1
2，
∴tan∠FAC=FH
AH=1
2，
∵FH=x2-2x-3，AH=x+1，
∴x2-2x-3
x+1=1
2，
X 1 = 7
2, X2 = - 1,
∴F（7
2，9
4）．
In the plane rectangular coordinate system xoy, the square of the parabola y = ax + BX + C intersects at two points a and B on the X axis (point a is on the left side of point B) and Y axis intersects point C. the coordinate of point a is (- 3,0). If the straight line y = KX + B passing through two points a and C is shifted down 3 units along the Y axis and passes through the origin exactly, and the symmetry axis of the parabola is a straight line x = - 2 Let △ BPP and △ BPP be the area of △ BPP, respectively ABP:S Δ BPC = 2:3, find the coordinates of point P
The first two questions of question 28 in Chengdu mathematics middle school entrance examination in 2010: the answers are as follows:
It is known that in the plane rectangular coordinate system xoy, the parabola y = AX2 + BX + C intersects the point C (0,4) with the Y axis, and intersects with the X axis at two points a and B, point a is on the left side of point B, Tan ∠ BCO = 1 And s △ AOC: s △ BOC = 4:1
In RT △ BOC
∵OC=4，tan∠BCO=1
Four
/ / ob = 1, so the coordinates of point B are (1, 0)
∵S△AOC：S△BOC=4：1
∴AO：OB=4：1
∵OB=1
/ / AO = 4, that is, the coordinates of point a are (- 4, 0)
Let the analytic formula of parabola be y = a (x + 4) (x-1)
Because the parabola passes through the coordinates of point C (0, 4), then there is
4×（-1）×a=4
∴a=-1
The analytic expression of the is parabola
y=-（x+4）（x-1）=-x2-3x+4．
In the plane rectangular coordinate system xoy, the parabola y = - 1 / 2x ^ 2 + BX + C passes through points a (1,3), B (0,1) Find the expression of parabola and vertex coordinates
This kind of question
Put the point into the parabola, C = 1, B = 5 / 2
Y = - 1 / 2x ^ 2 + 5x / 2 + 1 = - 1 / 2 (X-5 / 2) ^ 2 + 33 / 8, vertex (5 / 2,33 / 8)
As shown in the figure, in the plane rectangular coordinate system, it is known that the parabola y = ax 2 + Ba + C intersects the X axis on a (2,0), B (6,0) As shown in the figure, in the plane rectangular coordinate system, it is known that the parabola y = ax? + Ba + C intersects X axis at a (2,0), B (6,0), and intersects Y axis at point C (0,2 √ 3). (1) find the analytic formula of the parabola; (2) if the symmetry axis of the parabola intersects with the straight line y = 2x at point D, make the circle D tangent to the X axis, and intersect the y-axis with the two points of point E and F, find the length of inferior arc EF, (3) P is a point of parabola in the second quadrant image, PG is perpendicular to the x-axis, and the perpendicular foot is point g. try to determine the position of point P so that the area of △ PGA is divided into 1:2 parts by the line AC Hope to write the process that middle school students can understand
Y = ax 2 + Ba + C? Y = ax 2 + BX + C
(1) Substitute a (2,0), B (6,0) C (0,2 √ 3) into parabolic equation
4A + 2B + C = 0
36a+6b+c=0
c=2√3
A = √ 3 / 6 B = - (4 √ 3) / 3 C = 2 √ 3 is obtained
y=√3/6 x²-(4√3)/3x+2√3
(2) From the parabola equation
The axis of symmetry is x = 4
Substituting x = 4 into y = 2x, y = 8 is the radius
Draw a picture by yourself, you will find that the angle of the center of the circle to which the arc is directed is 120 degrees
So the arc length is 120 / 360 * 2 * 8 Π = 16 Π / 3
(3) Let PG be a straight line x = K
The linear equation of AC is y = - √ 3x + 2 √ 3
Substituting x = K into y = √ 3 / 6 x? - (4 √ 3) / 3x + 2 √ 3 and y = - √ 3x + 2 √ 3
y1=√3/6 k²-(4√3)/3k+2√3
y2=-√3k+2√3
y1=2y2
The solution is k = 0 or - 2
And because P is in the second quadrant, k = - 2
As shown in the figure, in the plane rectangular coordinate system, the symmetry axis of the parabola y = ax 2 + BX + C is a straight line x = - 3 / 2, and the intersection point between the parabola and the X axis is a and B, The point of intersection with Y-axis is C, the vertex of parabola is m, and the analytic formula of line MC is y = 3-4x-2 (1) Find the coordinates of vertex m (2) find the analytic formula of parabola (3) make circle P with the diameter of line AB, judge the position relationship between line MC and circle P, and prove your conclusion
(1) The vertex is on the axis of symmetry x = - 3 / 2
The analytic expression of MC is y = (3 / 4) x - 2
x = -3/2,y = -9/8 -2 = -25/8
M(-3/2,-25/8)
(2) y = ax²+bx+c = a[x + b/(2a)]²+ c -b^2/(4a)
The axis of symmetry is x = - B / (2a) = - 3 / 2, B = 3A (a)
C(0,-2)
-2 = 0 + 0 +c
c = -2 (b)
Vertex m ordinate C - B ^ 2 / (4a) = - 25 / 8 (c)
(a)(b)(c):a = 1/2,b = 3/2
The analytic formula of parabola: y = (1 / 2) x 2 + (3 / 2) x - 2
(3) y = (1/2)x² + (3/2)x - 2 = 0
(x+4)(x-1)= 0
A(-4,0),B(1,0)
Radius = (1 + 4) / 2 = 5 / 2
Center P (- 3 / 2,0)
The analytic formula of the line MC is y = (3 / 4) x - 2, 3x - 4Y - 8 = 0
The distance between the center of the circle and the line MC: | 3 (- 3 / 2) - 4 * 0 - 8 | / √ (3 | + 4 ﹤) = (25 / 2) / 5 = 5 / 2, which is equal to the radius, and the line MC is tangent to the circle
As shown in the figure, in the plane rectangular coordinate system, it is known that the parabola y = ax? + BX + C intersects the X axis at two points a (2,0) and B (6,0), and the intersection Y axis is at the point C (0,2 √ 3). (1) find the analytic formula of the parabola; (2) if the symmetry axis of the parabola and the straight line y = 2x intersect the point D, make the circle D tangent to the X axis, and the circle D intersects the Y axis at the two points E and F, and find the length of the inferior arc ef; (3) P is a point of the parabola in the second quadrant image, PG is perpendicular to the X axis, and the perpendicular foot is point g. try to determine the position of point P, so that the area of △ PGA is divided into two parts of 1:2 by the straight line AC. I hope to write the process that middle school students can understand. I don't want the answer!
(1) Let y = a (X-2) (y-6), and substitute the point C into the equation to get 3 = a (0-2) (0-6) under the root of 2, so a = 3 / 6 under the root sign
Y = (3 / 6 under radical) (X-2) (X-6)
(2) If the parabola intersects the x-axis at two points a and B, then the symmetry is x = 4, and the coordinate of point D is d (4,8)
The circle D is tangent to the X axis and its radius is 8. The equation of the circle is (x-4) ^ 2 + (Y-8) ^ 2 = 8 ^ 2 = 64
When x = 0, Y1 = 3 under 8-4, y2 = 3 under 8 + 4, e (3 under 0,8-4), f (3 under 0,8 + 4)
EF = 8 root sign 3
(3) Let AC cross PG at h,
If the area of △ PGA is divided into 1:2 parts by the line AC, then (pH * Ag) / (GH * Ag) = 1:2 or (GH * Ag /) (pH * Ag)) = 1:2
So GH = 2ph or pH = 2GH
Let g (x, 0), AC linear equation be y = - (3 under radical) (X-2)
1、GH=2PH
H (x, - (3 under radical) (X-2)), P (x, - 3 (under radical) (X-2) / 2)
If the point P is on a parabola, x = - 3, P (- 3,15 (3 under the radical) / 2)
2、PH=2GH
In the same way, x = - 12, P (- 12,42 (root 3))
As shown in the figure, in the plane rectangular coordinate system, the vertex coordinates of the parabola are known to be m (1,2), and pass through the point (0,3), and the parabola and the straight line x = 2 intersect at the point P (1) Solving the analytic formula of parabola (2) Take point a (2,5) on the line x = 2, and calculate the area of △ PAM (3) Whether there is a point Q on the parabola so that the area of △ QAM is equal to that of △ PAM, and the coordinates of point q are obtained This is mainly the third question,
(1)
Let the parabolic equation y = a (x + B / 2a) ^ 2 + (4ac-b ^ 2) / (4a)
X = 1, y = 2, x = 0, y = 3
-b/2a=1
(4ac-b^2)/4a=2
C=3
The result is a = 1 b = - 2 C = 3
The analytic formula of parabola is y = x ^ 2-2x + 3
(2)
Find the coordinates of point P: let x = 2 get y = 4-4 + 3 = 3P coordinates (2,3)
S△PAM=(5-3)*(2-1)/2=1
(3) In fact, the question is whether there is a point on the parabola and point P symmetrical about the line am
Slope of straight line am: (2-5) / (1-2) = 3
Slope of straight line PQ: - 1 / 3
Let PQ equation of straight line be: y = - X / 3 + B x = 2, y = 3
b=11/3
Y = - X / 3 + 11 / 3, find the intersection point of y = x ^ 2-2x + 3
-x/3+11/3=x^2-2x+3
Organize, get
3x^2-5x-2=0
(x-2)(3x+1)=0
X = 2 (round off) x = - 1 / 3, where y = 34 / 9
Conclusion: there exists this point Q, coordinates (- 1 / 3,34 / 9)
Draw the straight line x = - 1 and the straight line y = 2 in the plane rectangular coordinate system
In a rectangular plane coordinate system,
The line x = - 1 and the line y = 2 are shown in the following figure:
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