1. In the isosceles trapezoid ABCD, AB=DC=5, AD=4, BC= 10. Point E is on the bottom BC, and point F is on the waist AB.
(1) If EF bisects the circumference of the isosceles trapezoid ABCD, let the length of BE be x, and try to express the area of △BE with an algebraic expression containing x;
(2) Is there a line segment EF that bisects the circumference and area of the isosceles trapezoid ABCD? If yes, find out the length of BE at this time; If it does not exist, please explain the reason;
(3) Is there a line segment EF that divides the circumference and area of the isosceles trapezoid ABCD into 1∶2 at the same time? If yes, find out the length of BE at this time; If it does not exist, please explain why.
[Resolution] (1) comes from:
The trapezoid has a circumference of 12, a height of 4 and an area of 28.
FG⊥BC passes through point F at point G.
Pass point a as AK⊥BC in K.
Then: fg = 12-X5× 4.
∴S△BEF= 12, right? FG=-25 x2+245 x(7≤x≤ 10)
(2) existence
from( 1):-25 x2+245 x = 14。
Get x 1=7, x2=5 (don't give up).
∴ There is a line segment EF that bisects the circumference and area of the isosceles trapezoid ABCD at the same time. At this time BE=7.
(3) does not exist
Assuming it exists, it is obvious that: S△BEF∶SAFECD= 1∶2, (BE+BF)∶AF+AD+DC)= 1∶2.
Then -25x2+ 165x = 283.
Finishing: 3x2-24x+70 = 0
△= 576-840 & lt； 0
There is no such real number X.
That is, there is no perimeter and area of isosceles trapezoid ABCD with segment EF.
It is divided into two parts at the same time: 1∶2.
2. It is known that the intersection of parabola and Y axis is C, the vertex is M, and the analytical formula of the length of straight line CM and line segment CM is
(1) Find the analytical formula of parabola.
B(X2) Let the parabola and the X-axis have two intersections A (X 1, 0) and B (X2, 0), and point A is on the left side of B. Find the length of line segment AB.
(3) If AB is ⊙N, please judge the positional relationship between the straight line CM and ⊙N and explain the reasons.
[Analysis] (1) Solution 1: It is known that the straight line CM: y =-x+2 intersects with the Y axis at point c (0,2), and the parabola passes through point c (0,2), so c=2, and the vertex m of the parabola is on the straight line CM, so,
If b = 0, point C and point M coincide, which is irrelevant, so b =-2. Namely m
Make the vertical line of Y axis pass through point M, and the vertical foot is q, in.
So,, solution,.
The required parabola is: or the same as the following.
(1) solution 2: c (0,2) is obtained from the meaning of the question, and the coordinate of point m is M(x, y).
Point m is on a straight line.
From the Pythagorean Theorem
=, that is
solve an equation
∴ M' (2-2-2,4) or M' (2,0)
When m (-2,4), let the analytical expression of parabola be, ∫ the parabola passes through (0,2) point,
∴ ,∴
When m' (2,0), let the parabolic analytical formula be
The parabola of ∴. ∴ passes through the (0,2) point.
∴ The required parabola is: or
(2) The parabola has two intersections with the X axis,
∴ irrelevant, give it up.
∴ Parabola should be:
There are two intersections between parabola and X axis, and point A is on the left side of B.
(3)∫ab is the diameter of ∫n, ∴r =, n (-2,0), and ∫ M (-2,4), ∴MN = 4.
Let a straight line intersect the X axis at point D, then d (2 2,0), ∴DN = 4, MN = DN, ∴.
, such as NG⊥CM in g, in = R.
That is, the distance from the center of the circle to the straight line CM is equal to the radius ⊙ n.
∴ The straight line CM is tangent to⊙ n.
3. Known parabola
What is the value of (1)m, and there are two intersections between parabola and X axis?
(2) If the parabola and the X axis intersect at two points, M and N, when =3 and ≦, find the analytical expression of the parabola;
(3) If the vertex of the parabola in (2) is C, the intersection with the Y axis is above the origin, the symmetry axis of the parabola intersects with the X axis at point B, and the straight line y= -x+3 intersects with the X axis at point A ... Point P is the moving point on the symmetry axis of the parabola, the crossing point P is PD⊥AC, and the vertical foot D is on the straight line AC. Question: Is there a point P that makes? If it exists, find the coordinates of point P; If it does not exist, please explain why.
[Analysis] (1)∵ The parabola intersects the X axis at two points ∴△ > 0
In other words, the solution is: m
(2)∵ =3 ∴
When ∴m=2, m=-3.
∴
When ∴m=0, m=- 1
∴ when m=0, (conflict ≦, give up)
∴m=- 1
(3)∵ The intersection of parabola and Y axis is above the origin, ∴,
∴C(- 1,4),B(- 1,0)
The straight line y=-x+3 intersects the X axis at point A ∴ A (3 3,0).
∴BA=BC ∠PCD=45
When point d is on the AC line, let PD=DC=x,
∴ Solution:
When, Ⅷ
When, Ⅷ
When point d is on the extension line of AC, let PD=DC=x,
∴ Solution:
When, Ⅷ
When? (Give up)
When point d is on the extension line of CA, let PD=DC=x,
∴ Solution:
When, Ⅷ
When? (Give up)
∴ , , , 。
4. As shown in the figure, it is known that the image of parabola L 1: y=x2-4 intersects with X at points A and C,
(1) If parabola l2 and l 1 are axisymmetrical about X, find the analytical expression of l2; (3 points)
(2) If point B is the moving point on parabola l 1 (B does not coincide with A and C), the fourth vertex of the parallelogram with AC as the diagonal and A, B and C as the vertices is set to D, which proves that point D is on l2; (4 points)
(3) Inquiry: When point B is located on the image of l 1 in the upper and lower parts of the X axis, does the area of parallelogram ABCD have a maximum and a minimum? If it exists, judge what kind of special parallelogram it is and find its area; If it does not exist, please explain why. (4 points)
[Analysis] (1) Let the analytical formula of l2 be y = a (x-h) 2+k.
∵ The intersection of L2 and X axis A (-2,0), C (2,0), vertex coordinate (0,4), L 1 and L2 are symmetrical about X axis.
∴l2 passes through a (-2,0), c (2,0), and the vertex coordinate is (0,4).
∴y=ax2+4
∴0=4a+4 a=- 1
The analytical formula of ∴l2 is y=-x2+4.
(2) Let B(x 1, y 1).
Point b is on l 1
∴B(x 1，x 12-4)
∵ Quadrilateral ABCD is a parallelogram, and A and C are symmetrical about O.
∴B and d are symmetric about o
∴D(-x 1，-x 12+4)。
Substitute the coordinates of D(-x 1, -x 12+4) into L2: y =-x2+4.
Left = right
Point D is on l2.
(3) Let the area of parallelogram ABCD be s, then
S = 2 * S△ABC = AC * | y 1 | = 4 | y 1 |
A. when point b is above the x axis, y 1>0 > 0.
∴S=4y 1, which is a function proportional to s about y 1 and increases with the increase of y 1.
∴S has neither upper limit nor lower limit.
B when point b is below the x axis, -4 ≤ y 1 < 0.
∴S=-4y 1, which is a function about y 1 proportional to s, and decreases with the increase of y 1.
When y 1 =-4, the maximum value of s is 16, but it has no minimum value.
At this time, B(0, -4) is on the Y axis, and its symmetry point D is also on the Y axis.
∴AC⊥BD
∴ parallelogram ABCD is a diamond
At this time, the maximum S = 16.
5. As shown in figure 1, there are two right triangles ABC and EFG(A with the same shape (point A coincides with point E). It is known that AC = 8 cm, BC = 6 cm, ∠ C = 90, EG = 4 cm, ∠ EGF = 90, and O is the midpoint on the hypotenuse of △EFG.
As shown in Figure ②, if the whole △EFG starts from the position in Figure ① and moves in the direction of ray AB at the speed of 1cm/s, and △EFG moves, then point P starts from the vertex G of △EFG and moves to point F on the right-angled side GF at the speed of 1cm/s, and when point P reaches point F, it stops moving.
(1) When is the value of x, OP‖AC?
(2) Find the functional relationship between Y and X, and determine the range of independent variable X. 。
(3) Is there a moment of 13∶24 for the ratio of quadrilateral OAHP area to △ABC area? If it exists, find the value of x; If it does not exist, explain why.
(Reference data:1142 =12996,1152 =13225,1162 =/.
Or 4.42 = 19.36, 4.52 = 20.25, 4.62 = 2 1. 16).
[resolution] (1)∫Rt△EFG∽Rt△ABC
∴ , .
∴fg= = 3 cm.
When p is the midpoint of FG, OP‖EG, EG‖AC,
∴OP‖AC.
∴ x = = ×3= 1.5(s)。
∴, op ‖ ac when x is1.5s.
(2) In Rt△EFG, by Pythagorean theorem, EF =5cm. ..
∵EG‖ Ah,
∴△EFG∽△AFH。
∴ .
∴ .
∴ AH= ( x +5)，FH= (x+5)。
O is OD⊥FP, and the vertical foot is D.
Point o is the midpoint of EF,
∴OD= EG=2cm。
∫FP = 3-x，
∴S quadrilateral oahp = s △ afh-s △ ofp
= ? Huh? FH-？ OD？ freezing point
= ? (x+5)？ (x+5)- ×2×(3-x)
= x2+ x+3
(0