A diagonal of 1. parallelogram ABCD is fixed at two points A(3,-1) and C(2, -3), and point D moves along a straight line 3x-y+ 1=0, then the trajectory equation of point B is ().

A.3x-y-20=0 B.3x-y+ 10=0

C.3x-y-9=0 D.3x-y- 12=0

Answer: A solution: Let the midpoint of AC be O, that is, let the symmetry point of B(x, y) about O be (x0, y0), that is, D(x0, y0), then 3x-y-20=0.

2. If a tangent is drawn from a point on the straight line y=x+ 1 to the circle (x-3)2+y2= 1, the minimum tangent length is ().

A. 1

C.-2D.3

Answer: C Solution: When this point is the intersection of the vertical line drawn through the center of the circle and the straight line, the tangent length is the smallest. Because the distance from the center of the circle (3,0) to the straight line is d==2, the minimum tangent length is l==.

3. The straight line y=x+b and the curve x= have only one intersection point, so the value range of b is ().

A.{b||b|=}

B.{b|- 1

C.{ b |- 1≤b & lt； 1}

D. not the answer above

Answer:

B solution idea: in the same coordinate system, draw the image of y=x+b and curve x= (i.e. x2+y2= 1, x≥0), as shown in the figure, when tangent, b=- when other positions meet the conditions-1.

4. If the circle c: x2+y2+2x-4y+3 = 0 is symmetrical about the straight line 2ax+by+6=0, then the minimum tangent length from point (a, b) to the circle is ().

A.2 B.3

C.4 D.6

Answer: Solution C: The standard equation of a circle is (x+ 1)2+(y-2)2=2, so the center of the circle is (-1, 2) and the radius is. Because the circle is symmetrical about the line 2ax+by+6=0, the center of the circle is on the line 2ax+by+6=.

d==

==.

So when a=2, d has a minimum of 3 and the tangent length is the smallest, which is ==4, so c 。

5. It is known that the sum of the distances from the fixed point P to the two fixed points A and B is 8, and |AB|=4, and the midpoint of the line segment AB is O. Among all the line segments formed by the intersection of straight lines passing through point O and the trajectory of point P, the length of which is an integer is ().

A.5 Article B.6

C.7 Article D.8

Answer: Proposition D: This question examines the definition and nature of the ellipse, and the difficulty is moderate.

Solution: According to the meaning of the question, the trajectory of the moving point P is an ellipse with the focus on A and B, the length of the major axis is 8 and the length of the minor axis is 2=4. It should be noted that the shortest chord length passing through the ellipse center O is 4 and the longest chord length is 8, so the length of the line segment formed by the intersection of all straight lines passing through point O and the trajectory of point P can be integers 4, 5, 6, 7, 8, where the length is 4.

6. Let m, nR, if the straight line (m+ 1)x+(n+ 1)y-2=0 and the circle (x-1) 2+(y-1) 2 =1.

A.D.

Answer: D analysis: According to the topic, in OPQ, =, that is, |OP|≤2, and P(x0, x0-2), then x+(x0-2)2≤4, and the solution is x0, so D is chosen.

9. A straight line passing through point P( 1, 1) divides the circular area {(x, y)|x2+y2≤4} into two parts, so that the difference between the two parts becomes the equation of the straight line ().

A.x+y-2=0 B.y- 1=0

C.x-y=0 D.x+3y-4=0

Answer: Proposition A: It is moderately difficult to examine the comprehensive application of straight lines, linear programming and circles and the idea of combining numbers and shapes.

Solution: In order to divide the area of a circle into two parts with a straight line, the chord length of the circle passing through point P must be minimized, so the straight line should be perpendicular to the straight line OP. If point P( 1, 1) is known, then kOP= 1, so the slope of the straight line is-1. Similarly, a straight line through point p is needed.

10. The straight line y=kx+3 intersects with the circle (x-2)2+(y-3)2=4 at two points, m and n. If |MN|≥2, the value range of k is ().

A.B.

C.[-，] D。

Answer: Proposition B: It is moderately difficult to examine the positional relationship between a straight line and a circle.

Solution: In the right triangle composed of chord center distance d, radius r and half chord length |MN|, we can get |MN|=≥, 4-d2≥3, d2≤ 1, d==, k2≤, so -≤k≤ from Pythagorean theorem.

Second, fill in the blanks

1 1. It is known that the straight line L: Y =-(x- 1) and the circle O: X2+Y2 = 1 intersect at the point M in the first quadrant, and the L and Y axes intersect at the point A, then the area of MOA is equal to _ _ _ _ _ _.

Answer: Proposition: This question is less difficult, and the application of the positional relationship between a straight line and a circle is examined.

Thinking of solving problems: xM= can be obtained from simultaneous equations of straight lines and circles, so SMOA=×|OA|×xM=××=.

12. In ABC, the opposite sides of angles A, B and C are A, B and C respectively. If a2+b2=c2, the chord length of the straight line ax-by+c=0 cut by the circle x2+y2=9 is _ _ _ _ _.

Answer: 2 Proposition: This question examines the application of the positional relationship between a straight line and a circle. The chord length is generally solved by geometric method, which is less difficult.

Solution: The distance from the center of the circle to the straight line D = =, so the chord length of the straight line cut by the circle is 2=2=2.

13. given two points a (-2,0) and B( 1 0), the moving point p is not on the x axis and satisfies APO=BPO, where o is the origin, then the trajectory equation of point p is _ _ _ _ _.

Answer: (x-2)2+y2=4(y≠0) Proposition: It is moderately difficult to investigate the properties of the bisector and directly solve the trajectory equation.

Solution: Because A (-2,0) and B( 1 0) are two points, and the moving point P is not on the X axis and satisfies APO=BPO, so the point P is on the bisector of the angle APB, so use PAPB=AOOB=2 1 and set point P (x, y).

14. If the lengths of the segments cut by two parallel lines L 1: X-Y+ 1 = 0 and L2: X-Y+3 = 0 are 2, the inclination of m can be

15 30 45 60 75

The serial number of the correct answer is _ _ _ _ _. (Write the serial numbers of all the correct answers)

Answer: Solution: Let straight lines M and l 1, and l2 intersect at point A and point B respectively.

If a is ACl2 in c, then |AC|=.

|AB|=2，ABC=30。

The inclination angle of the straight line is 45 degrees,

The inclination of the straight line M is 45+30 = 75 or 45-30 = 15.

Group b

First, multiple choice questions

1. Given that the focus of parabola C: Y2 = 4x is f and the straight line y=2x-4 intersects with c at points A and B, then cos AFB= ().

A.B.

C.- D ...

Answer: D solution: x2-5x+4=0 means simultaneous elimination of Y, and the solution is x= 1 or x=4.

Let's set point A below the X axis, so A( 1, -2) and b (4,4).

Because F( 1, 0), so =(0, -2), =(3, 4).

So cos AFB=

= =-.So choose D.

2. It is known that there is a moving chord AB with a length of 6 on the parabola x2=4y, so the shortest distance from the midpoint of AB to the X axis is ().

A.B.

C. 1 D.2

Answer: D solution: According to the meaning of the question, the directrix L of the parabola is y=- 1, the passing A is AA 1 L in A1,the passing B is BB 1 L in B1,the midpoint of the chord AB is M, and the passing M is M65438. |AB|≤|AF|+|BF|(F is the focus of parabola), that is |AF|+|BF|≥6, that is | AA 1 |+BB 1 | ≥ 6, that is, 2|MM 1|≥6, |

3. Let hyperbola-=1(a >; 0, b>0) are F 1, F2, a is a point on the hyperbolic asymptote AF2F 1F2, and the distance from the origin o to the straight line AF 1 is |OF 1|, then the slope of the asymptote is ().

A. or -b or-

C. 1 or-1d. or-

Answer: Proposition D: This question examines the exploration of the geometric properties of hyperbola, which embodies the ingenious application of the mathematical thinking method of analytic geometry and is moderately difficult.

Solution: As shown in the figure, let's set point A as a point on the hyperbolic asymptote y=x in the first quadrant. The coordinates of point A can be obtained from AF2F 1F2, and from OBAF 1 and |OB|=|OF 1|, it can be obtained that sin is 1B=, and Tan is 65438.

4. Let F 1 and F2 be ellipses+=1(a >; B>0), the tangent F2 of the straight line y=b intersects the ellipse at point E, and E is just the tangent point of the straight line EF 1 and F2, then the eccentricity of the ellipse is ().

A.B.

C.D.

Answer: C solution idea: According to the meaning of the question, EF 1F2 is a right triangle, and F 1EF2 = 90.

|F 1F2|=2c，|EF2|=b，

According to the definition of ellipse |EF 1|=2a-b,

| ef 1 | 2+| ef2 | 2 = | f 1 F2 | 2，

That is, (2a-b)2+b2=(2c)2, b=a,

So e2===, so e=, so C.

5. The center of the equilateral hyperbola C is at the origin, the focus is on the X axis, the directrix of C and parabola y2= 16x intersects at point A and point B, |AB|=4, then the real axis length of C is ().

A.B.2 C.4 D.8

Answer: Solution C: According to the meaning of the question, let the equation of equilateral hyperbola be -= 1, and the directrix equation of parabola be y2= 16x be x=-4. If the equation is substituted into hyperbola, Y2 = 16-A2Y =+, then 2=4, and the solution is a=2, then the real axis length of hyperbola.

6. The area of the triangle surrounded by the directrix of parabola y2=- 12x and the two asymptotes of hyperbola -= 1 is equal to ().

A.B.3 C. D.3

Answer: Proposition B: This question mainly examines the basic knowledge such as the properties of parabola and hyperbola, aiming at examining the calculation ability of candidates.

Solution: According to the meaning of the question, the directrix equation of parabola y2=- 12x is x=3, the asymptote equation of hyperbola -= 1 is Y = X, and the coordinate of the intersection of straight line x=3 and straight line Y = X is (3,), so the area of triangle is equal to× 2× 3 = 3, so we choose it.

7. If hyperbola -= 1, ellipse+=1(m >; B>0) is greater than 1, then the triangle with sides of A, B and M must be ().

A. isosceles triangle B. right triangle

C. acute triangle D. obtuse triangle

Answer: Solution D: The eccentricity of hyperbola is e 1=, and the eccentricity of ellipse is e2=. We can know from the meaning of the question that E 1 E2 > 1, that is, b2(m2-a2-b2)>0, so m2-a2-b2 >; 0, namely m2 & gtA2+b2. According to the cosine theorem, the triangle is an obtuse triangle, so D is chosen.

8.F 1, F2 is a hyperbola-=1(a >; 0, b>0), the straight line L passing through F 1 intersects the left and right branches of hyperbola at point A and point B respectively. If ABF2 is an equilateral triangle, the eccentricity of hyperbola is ().

The second century BC.

Answer: Proposition B: This question mainly examines the definition, standard equation, geometric properties and calculation of basic quantities of hyperbola, and examines the reasoning ability and operational solving ability of candidates.

Solution: As shown in the figure, hyperbolic definition, | BF1|-BF2 | = | AF2 |-AF1| = 2A, because AF2 is a regular triangle, |BF2|=|AF2|=|AB|.

9. It is known that the minimum value of the sum of the distances between the moving point P on the straight line l 1: 4x-3y+6 = 0 and the straight line l2: x =- 1 and the parabola y2=4x to the straight line l 1 and the straight line l2 is ().

A.2 B.3

C.D.

Answer: A solution: Let the distances from the moving point P on the parabola y2=4x to the straight line l 1 and the straight line l2 be d 1 and d2, respectively. According to the definition of parabola, we can know that the straight line L2: X =- 1 is exactly the directrix of parabola, and the focus of parabola is F( 1, 0), then D2 = | PF.

10. Known hyperbola-=1(A > 0, b>0), A and B are two vertices of a hyperbola, P is a point on the hyperbola, and it is on the same branch of the hyperbola with point B, and the symmetry point of P about the Y axis is Q. If the slopes of straight lines AP and BQ are k 1, k2, K 1 K2 =- respectively, the eccentricity of the hyperbola is ().

A.B. C. D。

Answer: Proposition C: This question examines the solution of hyperbolic equation and its eccentricity, and examines the ability to simplify deformation, with moderate difficulty.

Thinking of solving problems: let A(0, -a), B(0, a), P(x 1, y 1), Q(-x 1, y 1), so k1k2.

Second, fill in the blanks

1 1. It is known that the focus of parabola y2=4x is f, and the straight line passing through point p (2,0) intersects with parabola at two points A(x 1, y 1) and B(x2, y2), then (1) y. (2) The minimum value of the triangular ABF area is _ _ _ _ _.

Answer: (1)-8 (2)2 Proposition: This question mainly investigates the positional relationship between a straight line and a parabola, with moderate difficulty.

Solution: Let the equation of straight line AB be x-2=m(y-0), that is, x=my+2, and get y2-4my-8=0. (1) According to the relationship between root and coefficient, y 1y2=-8. (2) The area of triangle ABF is s = | fp.

Knowledge expansion: solving ABF after segmentation can effectively reduce the amount of calculation.

12.B 1, B2 is the two ends of the short axis of the ellipse, o is the center of the ellipse, and the vertical line passing through the left focus F 1 as the long axis intersects the ellipse at p, if |F 1B2| equals | of 1b2 |.

Answer: Proposition conception: This question examines the basic properties of ellipse and the properties of the equal ratio term, and the difficulty is moderate.

Solution: Let the elliptic equation be+=1(A > B>0), let x=-c, and get y2=, |PF 1|=. = =, then | f 1b2 | | b 1b2 |, and a2 = 2bc.a4 = 4b2 (.

13. Known parabola c: y2 = 2px (p > The directrix of 0) is L, and the straight line with slope passing through M( 1, 0) intersects with L at point A, and the intersection with C is B. If =, then P = _ _ _ _ _ _

Answer: 2 solution idea: make b perpendicular to the directrix l and e,

=, m is the midpoint of AB,

|BM|=|AB|, with a slope of,

BAE=30，|BE|=|AB|，

|BM|=|BE|, m is the focus of parabola,

p=2。

14.

As shown in the figure, the center of the ellipse is at the coordinate origin O, the vertices are A 1, A2, B 1, B2, and the focus is F 1, F2. The extension line B 1F2 intersects with A2B2 at point p. If B 1PA2 is obtuse, the eccentricity range of this ellipse is _ _.

Answer: solution idea: let the equation of ellipse be+=1(a > B>0), B 1PA2 is an obtuse angle, which can be converted into (a, -b) (-c, -b)0, e >；; Or e