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sat数学单词你们知道〔sat〕数学考试是中国的考生的强势科目，但是前提是看得懂单词。
下面是我为大家整理的关于sat数学单词的相关资料，希望帮到大家。
sat数学单词之代数algebraic term 代数项like terms，similar terms 同类项numerical coefficient 数字系数literal coefficient 字母系数inequality 不等式triangle inequality 三角不等式range 值域original equation 原方程linear equation with one unknown 一元一次等式linear inequality with one unknown 一元一次不等式equivalent equation 同解方程linear equation 线性方程sat数学单词之函数：function 函数coordinate system 坐标系rectangular coordinate 直角坐标系origin 原点abscissa 横坐标ordinate 纵坐标numberline 数轴quadrant 象限slope 斜率complex plane 复平面linear function 一次函数inverse proportional function 反比例函数 quadratic function 二次函数parabola 抛物线proportional function 正比例函数exponential function 指数函数logarithmic function 对数函数odd function 奇函数even function 偶函数intersect 相交vertical / perpendicular 垂直parallel 平行inverse function 反函数complementary function 余函数sat数学单词之统计学：verage平均数median中数mode众数(在一系列数中出现多的数) arithmetic mean算术平均数weighted average加权平均数geometric mean几何平均数maximum大值minimum小值range值域(一系列数中大值减小值) dispersion差量，离差standard deviation标准方差distribution(频数或频率)分布frequency distribution频数分布normal distribution正态分布sat数学单词之运算：add/plus加subtract/minus减multiply/times乘divide除total1总数(用在加法中，相当于+);2总计(用于减法中，相当于-)division1除;2部分divisible可被整除的divided evenly被整除dividend被除数divisor除数以上就是sat数学单词的内容，希望对大家有所帮助哦。
SAT考试数学练习题第九套 SAT problem solving practice test 91. (3 x 104) + (2 x 10²) + (4 x 10) =A. 302400B. 32400C. 30240D. 3240E. 3242. Andy solves problems 74 to 125 inclusive in a Math exercise. How many problems does he solve?A. 53B. 52C. 51D. 50E. 493.If x and y are integers, and 3x + 2y = 13, which of the following could be the value of y ?A. 0B. 1C. 2D. 3E. 44.In triangle ABC, AD = DB , DE is parallel to BC, and the area of triangle ABC is 40. What is the area of triangle ADE ?A. 10B. 15C. 20D. 30E. it cannot be determined from the information given5. If n > 0 , which of the following must be true?I n² > 1II n - n² < 0III 2n - 1 > 0A. I onlyB. II onlyC. III onlyD. I and II onlyE. none6.If the slope of a line is ½ and the y-intercept is 3, what is the x-intercept of the same line?A. 6B. 3/2C. 0D. -2/3E. -67.6 people meet for a business lunch. Each person shakes hands once with each other person present. How many handshakes take place?A. 30B. 21C. 18D. 15E. 108.If x² - y² = 55, and x - y = 11, then y =A. 8B. 5C. 3D. -8E. -39.In a sports club with 30 members, 17 play badminton and 19 play tennis and 2 do not play either. How many members play both badminton and tennis?A. 7B. 8C. 9D. 10E. 1110. Rectangle ABCD has a perimeter of 26. The half circle with diameter AD has an area of 8π. What is the perimeter of the part of the figure that is not shaded?A. 26 + 4πB. 18 + 8πC. 18 + 4πD. 14 + 4πE. 14 + 2πSAT数学练习题第9套参考答案1.Correct Answer: CExplanation:2 x 104 = 30,000; 2 x 102 = 200; 4 x 10 = 40The total is 30,2402.Correct Answer: BExplanation:To find how many problems in the series we need to take the difference and add one.125 - 74 = 51; 51 + 1 =523.Correct Answer: CExplanation:Substitute the given values for y and check whether you get an integer value for x.For example, using 0 we get 3x = 13; x = 13/3 which is not a whole number. The right answer is 2, since 3x + 2(2) = 13; 3x = 13 - 4 = 9; x = 9/3 = 3.4.Correct Answer: AExplanation:The big triangle ABC is similar to the small triangle ADE because their bases are parallel. If corresponding side of two similar triangles are known the ratio of the areas is also known. In this case, let AD be one unit, then AB is 2 units (given that AB = AD + DB). The ratios of the sides is 1 : 2. The ratio of the areas will be (1)2 : (2)2 ; 1 : 4Since the big triangle has area 40, using the ratio, the small has area 10.5.Correct Answer: EExplanation:Given that n is positive, it could be a positive fraction, 1, or a fraction of whole number greater than 1.If n = 1, then case I is not true since n2 = 1Likewise in II if n = 1, n - n2 = 0, and the statement is not true.In III, if n = ½, then 2n - 1 = 0, and again the statement is incorrect.6.Correct Answer: EExplanation:The equation for a straight line is y = mx + c, where m = slope and c = y-intercept.Putting the given values in this equation we have y = x/2 + 3The x-intercept occurs where y = 0. Thus, 0 = x/2 + 3 ; -3 = x/2 ; -6 =x7.Correct Answer: DExplanation:Imagine the first person of the six. He or she will have to shake hands with each of the other 5. Now turn to the second person. He or she will have to shake with the other five, but he she has already shaken with the first person. This means 4 new handshakes. The third person will have to shake with 5 - 2 = 3 people, and so on. Total handshakes = 5 + 4 + 3 + 2 + 1 = 158.Correct Answer: EExplanation:x2 - y2 can be expressed as (x + y)(x - y); since x - y =11 we can write (x + y)11 = 55; therefore x + y = 5Adding the two equations x + y = 5 and x - y = 11 we get2x = 16; x = 8Therefore 8 - y = 11; y = -39.Correct Answer: BExplanation:Since 2 do not play either, there are 28 members who play one sport or the other. Let the number who play both be n.Total (28) will be made up of only badminton players (17 - n), plus onlytennis (19 - n) and those who play both (n).28 = (17 - n) + (19 - n) + n28 = 36 - n ; n = 810.Correct Answer: CExplanation:The total perimeter of the un-shaded part is made up of three sides of the rectangle and the perimeter of the half-circle.The area of a half circle = ½ π r28 π = ½ π r2; therefore r = 4The perimeter of the half circle is ½ 8 π = 4 πThe diameter of the circle (8) = the length of the rectangle.Total perimeter of the rectangle = 26Three sides measure 26 - 8 = 18 Ans. 18 + 4π。
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Solving linear equations1. Solve for k:k + 22 = 29k =Correct answer: 7 Difficulty level: 12. Solve for n:-8 + n = 23n =Correct answer: 31 Difficulty level: 13. Solve for x:x – 9 = 1x =Correct answer: 10 Difficulty level: 14. Solve for n:18 = n - 18n =Correct answer: 36 Difficulty level: 15. Solve for k:k + 10 = 27k =Correct answer: 17 Difficulty level: 16. Solve for t:t + 25 = 26t =7. Solve for k:-30 + k = 22k =Correct answer: 52 Difficulty level: 18.Solve for k:19 = 14 + kk =Correct answer: 5 Difficulty level: 19. Solve for q:3 = -8 + qq =Correct answer: 11 Difficulty level: 110. Solve for p:10 = -19 + pp =Correct answer: 29 Difficulty level: 111. Solve for r:-12 + r = 7r =Correct answer: 19 Difficulty level: 112. Solve for p:p + 12 = 30p =Correct answer: 18 Difficulty level: 113. Solve for p:p – 18 = 3p =Correct answer: 21 Difficulty level: 114. Solve for k:30 = k + 23k =15. Solve for y:25 = -14 + yy =Correct answer: 39 Difficulty level: 116. Solve for r:20 = r + 11r =Correct answer: 9 Difficulty level: 117. Solve for z:-26 + z = 15z =Correct answer: 41 Difficulty level: 118. Solve for p:29 = p + 20p =Correct answer: 9 Difficulty level: 119. Solve for y:24 = y + 19y =Correct answer: 5 Difficulty level: 120. Solve for n:27 = 15 + nn =Correct answer: 12 Difficulty level: 121. Solve for k:2 = -30 + kk =Correct answer: 32 Difficulty level: 122. Solve for n:19 + n = 30n =。
SAT 数学知识点一Number and Operations Review 一、Properties of integers知道下列说法表示的内容：1. Integers consist of the whole numbers and their negatives (including zero).2. Integers extend infinitely in both negative and positive directions.3. Integers do not include fractions or decimals.4. Negative integers5. Positive integers6. The integer zero is neither positive nor negative.7. odd numbers（奇数）and even numbers（偶数）8. Consecutive integers9. Addition of integers(奇数偶数的加法规则)10. Multiplication of integers（奇数偶数的乘法规则）二、Arithmetic word problems（算术题）三、Number lines（数轴）四、Square and square roots（平方和平方根）五、Fractions and rational numbers（分数与有理数）六、Elementary number theory☆Factors, multiples, and remainders☆Prime numbers七、Ratios, proportions, and percents八、Sequences九、Sets(union, intersection, elements)十、Counting problems Counting problems involve figuring out how many ways you can select or arrange members of groups, such as letters of the alphabet, numbers or menu selections.☆Fundamental counting problems分步完成事件和分类完成事件发生的可能性☆Permutations and combinations （排列组合）基本排列组合理论十一、Logical reasoningThe SAT doesn’t include1.Tedious or long computations2.Matrix operations1.2.3.4.5. 6.7.8.9.10.11.12.13. 14.SAT数学知识点二Algebra and Functions Review Many math questions require knowledge of algebra. This chapter gives you some further practice. You have to manipulate and solve a simple equation for an unknown, simplify and evaluate algebraic expressions, and use algebraic expressions, and use algebraic concepts in problem-solving situations.For the math questions covering algebra and functions content, you should be familiar with all of the following basic skills and topics:一、Operations on algebraic expressions二、Factoring三、Exponents四、Evaluating expressions with exponents and roots五、Solving equations☆Working with “unsolvable” equations☆Solving for one variable in terms of another☆Solving equations involving radical expressions六、Absolute value 七、Direct translation into mathematical expressions八、Inequalities九、Systems of linear equations and inequalities十、Solving quadratic equations by factoring 十一、Rational equations and inequalities 十二、Direct and inverse variation十三、Word problems十四、Functions☆Function notation and evaluation☆Domain and range☆Using new definitions☆Functions as models☆Linear functions: their equations and graphs☆Quadratic functions: their equations and graphs☆Qualitative behavior of graphs and functions☆Translations and their effects on graphsand functionsThe SAT doesn’t include:一、Solving quadratic equations thatrequire the use of the quadraticformula二、Complex numbers三、Logarithms1.2.3.4. 5.6.7.8.9.10.SAT 数学知识点三Geometry and Measurement Review Concept you should to knowFor the mathematics questions covering geometry and measurement concepts, you should be familiar with all of the following basic skills, topics, and formulas:一、Geometric notation二、Points and lines三、Angles in the plane四、Triangles(including special triangles)☆Equilateral triangles☆Isosceles triangles☆Right triangles and the Pythagorean theorem ☆30º-60º-90ºtriangles☆45º-45º-90ºtriangles☆3-4-5 triangles☆Congruent triangles☆Similar triangles☆The triangle inequality五、Quadrilaterals☆Parallelograms☆Rectangles☆Squares六、Areas and Perimeters☆Areas of squares and rectangles☆Perimeters of squares and rectangles☆Area of triangles☆Area of Parallelograms七、Other polygons☆Angles in a polygon☆Perimeter☆Area八、Circles☆Diameter☆Radius☆Arc☆Tangent to a circle☆Circumference☆Area九、Solid geometry☆Solid figures and volumes☆Surface area十、Geometric perception十一、Coordinate geometry☆Slopes, parallel lines, and perpendicular lines☆The midpoint formula☆The distance formula十二、TransformationsThe SAT doesn’t include:一、Formal geometric proofs二、Trigonometry三、Radian measure1.2.3.4.5.6. 7.8.9.SAT 数学知识点四Data Analysis, Statistics andProbability ReviewFor the math questions covering data analysis, statistics and probability concepts, you should be familiar with all of the following basic skills and topics:一、Data interpretation二、Statistics☆Arithmetic mean☆Median☆Mode☆Weighted average☆Average of algebraic expression☆Using average to find missing numbers三、Elementary probability四、Geometric probabilityThe SAT doesn’t include:四、Computation of standard deviation 1.2.3.4.5. 6.7.8.Word Problems1.2.3.4. 5-75.6.7.1112。
SAT数学知识点（一）考试，如果一旦知道考点，对于考生就有一种如释重负的感觉。
所以，面对SAT数学，文都国际教育小编将给大家分享一下基本的SAT数学知识点，希望对大家的考试有所帮助。
代数部分（heart of Algebra）：核心是线性方程（组）、线性函数、线性不等式。
1.建立、解答或解释单变量的线性方程（create, solve, or interpret linear equations of one variable）学生根据题目背景（context）建立、解答和解释单变量（variable）的线性(linear)表达式（expression）或方程(equation)。
表达式或方程有理因数（rational coefficients）。
学生可能需要多个步骤简化表达式、简化方程或解出方程中的变量。
2.建立、解答或解释单变量的线性不等式（create, solve, or interpret linear inequalities of one variable）学生根据题目背景（context）建立、解答和解释单变量（variable）的线性不等式（linear inequalities）。
含有理分式的不等式（rational coefficients）。
学生可能需要多个步骤简化表达式、简化方程或解出方程中的变量的值。
3.根据两个变量的线性关系建立一个线性函数（build a linear function that models a linear relationship between two quantities）学生使用两个变量的方程（function）或函数记号（function notation）描述一个背景下的线性关系（linear relationship）。
方程（equation）或有理因式函数(function with rational coefficients）。
2024 SAT考试数学历年题目大盘点2024年即将迎来 SAT 考试，作为一项重要的学术考试，数学部分一直是考生们关注的焦点。
为了帮助考生更好地备考，本文将对近年的 SAT 数学试题进行盘点，并分析其中的一些常见考点。
一、代数与函数（Algebra and Functions）1. 方程与不等式（Equations and Inequalities）近年的 SAT 数学试题中，涉及到方程与不等式的题目较为常见。
考生需要熟练掌握一元一次方程、二元一次方程、一元二次方程以及绝对值不等式等的解法。
此外，还需要注意对于方程和不等式解集的理解和应用。
2. 函数（Functions）函数是 SAT 考试中的重点内容之一。
考生需要了解各类常见函数的性质，如线性函数、二次函数、指数函数、对数函数等，并能够灵活运用函数的性质解决问题。
此外，函数的符号表示、定义域、值域以及函数图像的理解也是考点之一。
二、几何（Geometry）1. 平面几何（Plane Geometry）几何部分涉及到平面几何和空间几何的知识。
在平面几何方面，考生需要掌握直线与角、三角形、四边形、圆等图形的性质，能够灵活运用相关定理解决几何问题。
2. 空间几何（Solid Geometry）空间几何侧重于三维几何图形的性质和计算。
考生需要了解球体、圆柱体、锥体、棱柱等立体图形的特征和计算方法，并能够运用相关定理解决与空间几何有关的问题。
三、数据分析（Data Analysis）1. 统计与概率（Statistics and Probability）数据分析部分主要考察考生对统计与概率的理解和运用能力。
考生需要了解统计数据的收集、整理和分析方法，包括频率分布、平均数、中位数、标准差等；同时还需要对概率的计算和应用有一定的掌握。
2. 数据表示与解释（Data Representation and Interpretation）数据表示与解释主要考察考生对数据图表的理解和分析能力。
unit九知识点总结Unit 9 is an important unit that covers a wide range of knowledge points related to the subject matter. It is essential for students to master the key concepts in this unit to be successful in their studies. In this summary, we will go over the main knowledge points in Unit 9 and provide a comprehensive overview of each topic.1. Algebraic ExpressionsAlgebraic expressions are mathematical expressions that contain variables, numbers, and mathematical operations. In this unit, students will learn how to simplify and evaluate algebraic expressions. They will also become familiar with the concept of like terms and how to identify and combine them.2. Solving EquationsSolving equations is a fundamental skill in mathematics. In this unit, students will learn various methods for solving linear equations, including the use of the addition and multiplication properties of equality. They will also explore the concept of equivalent equations and how to use this concept to solve problems.3. InequalitiesInequalities are mathematical expressions that compare the relative size of two quantities. Students in this unit will learn how to solve and graph inequalities on a number line. They will also study the concepts of compound inequalities and how to solve them using logical reasoning.4. Systems of EquationsA system of equations is a set of two or more equations that contain the same variables. In this unit, students will learn how to solve systems of linear equations using the substitution method, the elimination method, and graphing. They will also explore the concept of consistent and inconsistent systems and how to determine their solutions.5. PolynomialsPolynomials are algebraic expressions that contain one or more terms. Students will study the properties of polynomials, including their degree, leading coefficient, and constant term. They will also learn how to add, subtract, multiply, and factor polynomials.6. FactoringFactoring is the process of finding the factors of a polynomial. In this unit, students will learn various factoring techniques, including factoring by the greatest common factor, factoring trinomials, and factoring the difference of squares. They will also study theconcept of prime and composite numbers and how to use these concepts to factor a polynomial.7. Rational ExpressionsA rational expression is the ratio of two polynomials. Students in this unit will learn how to simplify, multiply, divide, add, and subtract rational expressions. They will also study the concept of restrictions on the domain of a rational expression and how to simplify complex fractions.8. Quadratic EquationsQuadratic equations are polynomial equations of the second degree. In this unit, students will learn how to solve quadratic equations using the square root property, factoring, completing the square, and the quadratic formula. They will also explore the concept of the discriminant and how to use it to determine the nature of the solutions of a quadratic equation.9. Exponents and RadicalsExponents and radicals are essential concepts in algebra. In this unit, students will study the properties of exponents and how to simplify expressions with exponents. They will also learn how to simplify and manipulate radical expressions, including rationalizing the denominator and solving radical equations.10. FunctionsA function is a relation between a set of inputs and a set of outputs. Students in this unit will learn how to evaluate functions, determine the domain and range of a function, and identify the symmetry of a function. They will also study the concepts of one-to-one and onto functions and how to determine if a function has an inverse.In summary, Unit 9 covers a wide range of important knowledge points related to algebra and functions. It is crucial for students to have a solid understanding of these concepts to succeed in their studies. By mastering the key concepts in this unit, students will be well-prepared to tackle more advanced topics in mathematics.。
SAT数学方程式和不等式知识考点对于SAT数学，我们想要SAT数学提分，其中一个很重要的就是牢固的掌握基础知识，只有在不必犯错的地方不丢分，然后才能更好的冲刺高分。
接下来文都国际教育小编就对在SAT数学的题目中会出现的方程式和不等式谈一谈，希望以下的知识点对同学们有所帮助。
在新SAT数学考查能力之一，Heart of Algebra, 代数能力，主要考察方程式和不等式，出现在Math部分。
因此我们首先来看一下SAT数学方程式和不等式的常用表达方式和要点。
Equations方程式Inequalities不等式Linear quations 直线方程式Word problem 数字问题System of equations方程式组规律：Doing the same thing to each side of an equation does not mean doing the same thing to each term of the equation.对于不等式的每一遍做同样的改动，不意味着对每一项都进行同样的改动对于SAT数学不少同学比较疑惑的一点就是英语单词和数学符号之间的关系，因此我们在SAT数学提分的时候，也需要弄懂方程不等式里面加减乘除会对应哪些英文单词。
下面同样我们为大家总结一下，在SAT数学题目中方程式和不等式中重要的数学符号表达方式：下面在对新SAT样题方程式和不等式一些常见考点，希望同学们在备考的时候要多加注意。
1、代数核心知识识别反映现实生活场景的不等式中的数学记号是否正确。
2、代数核心知识将已知的复合不等式变换形式，并找到一个满足所有条件的值。
3、代数核心知识将已知的复合不等式变换形式，并找到一个满足所有条件的值。
4、解决问题和分析数据理解一个反应现实情境的表达式或方程式，并能根据情景对全部或部分表达式进行解释。
5、代数核心知识建一个反映现实情境的方程。
6、代数核心知识建一个反映现实情境的线性方程组。
江西省南昌市2015-2016学年度第一学期期末试卷（江西师大附中使用）高三理科数学分析一、整体解读试卷紧扣教材和考试说明，从考生熟悉的基础知识入手，多角度、多层次地考查了学生的数学理性思维能力及对数学本质的理解能力，立足基础，先易后难，难易适中，强调应用，不偏不怪，达到了“考基础、考能力、考素质”的目标。
试卷所涉及的知识内容都在考试大纲的范围内，几乎覆盖了高中所学知识的全部重要内容，体现了“重点知识重点考查”的原则。
1．回归教材，注重基础试卷遵循了考查基础知识为主体的原则，尤其是考试说明中的大部分知识点均有涉及，其中应用题与抗战胜利70周年为背景，把爱国主义教育渗透到试题当中，使学生感受到了数学的育才价值，所有这些题目的设计都回归教材和中学教学实际，操作性强。
2．适当设置题目难度与区分度选择题第12题和填空题第16题以及解答题的第21题，都是综合性问题，难度较大，学生不仅要有较强的分析问题和解决问题的能力，以及扎实深厚的数学基本功，而且还要掌握必须的数学思想与方法，否则在有限的时间内，很难完成。
3．布局合理，考查全面，着重数学方法和数学思想的考察在选择题，填空题，解答题和三选一问题中，试卷均对高中数学中的重点内容进行了反复考查。
包括函数，三角函数，数列、立体几何、概率统计、解析几何、导数等几大版块问题。
这些问题都是以知识为载体，立意于能力，让数学思想方法和数学思维方式贯穿于整个试题的解答过程之中。
二、亮点试题分析1．【试卷原题】11.已知,,A B C 是单位圆上互不相同的三点，且满足AB AC →→=，则AB AC →→⋅的最小值为（ ）A ．14-B ．12-C ．34-D ．1-【考查方向】本题主要考查了平面向量的线性运算及向量的数量积等知识，是向量与三角的典型综合题。
解法较多，属于较难题，得分率较低。
【易错点】1．不能正确用OA ，OB ，OC 表示其它向量。
2．找不出OB 与OA 的夹角和OB 与OC 的夹角的倍数关系。
【解题思路】1．把向量用OA ，OB ，OC 表示出来。
2．把求最值问题转化为三角函数的最值求解。
【解析】设单位圆的圆心为O ，由AB AC →→=得，22()()OB OA OC OA -=-，因为1OA OB OC ===，所以有，OB OA OC OA ⋅=⋅则()()AB AC OB OA OC OA ⋅=-⋅-2OB OC OB OA OA OC OA =⋅-⋅-⋅+ 21OB OC OB OA =⋅-⋅+设OB 与OA 的夹角为α，则OB 与OC 的夹角为2α所以，cos 22cos 1AB AC αα⋅=-+2112(cos )22α=--即，AB AC ⋅的最小值为12-，故选B 。
【举一反三】【相似较难试题】【2015高考天津，理14】在等腰梯形ABCD 中,已知//,2,1,60AB DC AB BC ABC ==∠= ,动点E 和F 分别在线段BC 和DC 上,且,1,,9BE BC DF DC λλ==则AE AF ⋅的最小值为 .【试题分析】本题主要考查向量的几何运算、向量的数量积与基本不等式.运用向量的几何运算求,AE AF ，体现了数形结合的基本思想，再运用向量数量积的定义计算AE AF ⋅，体现了数学定义的运用，再利用基本不等式求最小值，体现了数学知识的综合应用能力.是思维能力与计算能力的综合体现. 【答案】2918【解析】因为1,9DF DC λ=12DC AB =，119199918CF DF DC DC DC DC AB λλλλλ--=-=-==， AE AB BE AB BC λ=+=+，19191818AF AB BC CF AB BC AB AB BC λλλλ-+=++=++=+，()221919191181818AE AF AB BC AB BC AB BC AB BCλλλλλλλλλ+++⎛⎫⎛⎫⋅=+⋅+=+++⋅⋅ ⎪ ⎪⎝⎭⎝⎭19199421cos1201818λλλλ++=⨯++⨯⨯⨯︒2117172992181818λλ=++≥+= 当且仅当2192λλ=即23λ=时AE AF ⋅的最小值为2918. 2．【试卷原题】20. （本小题满分12分）已知抛物线C 的焦点()1,0F ，其准线与x 轴的交点为K ，过点K 的直线l 与C 交于,A B 两点，点A 关于x 轴的对称点为D ． （Ⅰ）证明：点F 在直线BD 上； （Ⅱ）设89FA FB →→⋅=，求BDK ∆内切圆M 的方程. 【考查方向】本题主要考查抛物线的标准方程和性质，直线与抛物线的位置关系，圆的标准方程，韦达定理，点到直线距离公式等知识，考查了解析几何设而不求和化归与转化的数学思想方法，是直线与圆锥曲线的综合问题，属于较难题。
【易错点】1．设直线l 的方程为(1)y m x =+，致使解法不严密。
2．不能正确运用韦达定理，设而不求，使得运算繁琐，最后得不到正确答案。
【解题思路】1．设出点的坐标，列出方程。
2．利用韦达定理，设而不求，简化运算过程。
3．根据圆的性质，巧用点到直线的距离公式求解。
【解析】（Ⅰ）由题可知()1,0K -，抛物线的方程为24y x =则可设直线l 的方程为1x my =-，()()()112211,,,,,A x y B x y D x y -，故214x my y x =-⎧⎨=⎩整理得2440y my -+=，故121244y y m y y +=⎧⎨=⎩则直线BD 的方程为()212221y y y y x x x x +-=--即2222144y y y x y y ⎛⎫-=- ⎪-⎝⎭令0y =，得1214y yx ==，所以()1,0F 在直线BD 上.（Ⅱ）由（Ⅰ）可知121244y y m y y +=⎧⎨=⎩，所以()()212121142x x my my m +=-+-=-，()()1211111x x my my =--= 又()111,FA x y →=-，()221,FB x y →=-故()()()21212121211584FA FB x x y y x x x x m →→⋅=--+=-++=-，则28484,93m m -=∴=±，故直线l 的方程为3430x y ++=或3430x y -+=213y y -===±，故直线BD 的方程330x -=或330x -=，又KF 为BKD ∠的平分线，故可设圆心()(),011M t t -<<，(),0M t 到直线l 及BD 的距离分别为3131,54t t +--------------10分 由313154t t +-=得19t =或9t =（舍去）.故圆M 的半径为31253t r +== 所以圆M 的方程为221499x y ⎛⎫-+= ⎪⎝⎭【举一反三】【相似较难试题】【2014高考全国，22】 已知抛物线C ：y 2＝2px(p>0)的焦点为F ，直线y ＝4与y 轴的交点为P ，与C 的交点为Q ，且|QF|＝54|PQ|.（1）求C 的方程；（2）过F 的直线l 与C 相交于A ，B 两点，若AB 的垂直平分线l′与C 相交于M ，N 两点，且A ，M ，B ，N 四点在同一圆上，求l 的方程．【试题分析】本题主要考查求抛物线的标准方程,直线和圆锥曲线的位置关系的应用,韦达定理,弦长公式的应用,解法及所涉及的知识和上题基本相同. 【答案】（1）y 2＝4x. （2）x －y －1＝0或x ＋y －1＝0. 【解析】（1）设Q(x 0，4)，代入y 2＝2px ，得x 0＝8p，所以|PQ|＝8p ，|QF|＝p 2＋x 0＝p 2＋8p.由题设得p 2＋8p ＝54×8p ，解得p ＝－2(舍去)或p ＝2，所以C 的方程为y 2＝4x.（2）依题意知l 与坐标轴不垂直，故可设l 的方程为x ＝my ＋1(m≠0)． 代入y 2＝4x ，得y 2－4my －4＝0. 设A(x 1，y 1)，B(x 2，y 2)， 则y 1＋y 2＝4m ，y 1y 2＝－4.故线段的AB 的中点为D(2m 2＋1，2m)， |AB|＝m 2＋1|y 1－y 2|＝4(m 2＋1)．又直线l ′的斜率为－m ，所以l ′的方程为x ＝－1m y ＋2m 2＋3.将上式代入y 2＝4x ，并整理得y 2＋4m y －4(2m 2＋3)＝0.设M(x 3，y 3)，N(x 4，y 4)，则y 3＋y 4＝－4m，y 3y 4＝－4(2m 2＋3)．故线段MN 的中点为E ⎝ ⎛⎭⎪⎫2m2＋2m 2＋3，－2m ，|MN|＝1＋1m 2|y 3－y 4|＝4（m 2＋1）2m 2＋1m 2.由于线段MN 垂直平分线段AB ，故A ，M ，B ，N 四点在同一圆上等价于|AE|＝|BE|＝12|MN|，从而14|AB|2＋|DE|2＝14|MN|2，即 4(m 2＋1)2＋⎝ ⎛⎭⎪⎫2m ＋2m 2＋⎝ ⎛⎭⎪⎫2m 2＋22＝4（m 2＋1）2（2m 2＋1）m 4，化简得m 2－1＝0，解得m ＝1或m ＝－1， 故所求直线l 的方程为x －y －1＝0或x ＋y －1＝0.三、考卷比较本试卷新课标全国卷Ⅰ相比较，基本相似，具体表现在以下方面： 1. 对学生的考查要求上完全一致。
即在考查基础知识的同时，注重考查能力的原则，确立以能力立意命题的指导思想，将知识、能力和素质融为一体，全面检测考生的数学素养，既考查了考生对中学数学的基础知识、基本技能的掌握程度，又考查了对数学思想方法和数学本质的理解水平，符合考试大纲所提倡的“高考应有较高的信度、效度、必要的区分度和适当的难度”的原则． 2. 试题结构形式大体相同，即选择题12个，每题5分，填空题4 个，每题5分，解答题8个（必做题5个），其中第22，23，24题是三选一题。
题型分值完全一样。
选择题、填空题考查了复数、三角函数、简易逻辑、概率、解析几何、向量、框图、二项式定理、线性规划等知识点，大部分属于常规题型，是学生在平时训练中常见的类型．解答题中仍涵盖了数列，三角函数，立体何，解析几何，导数等重点内容。
3. 在考查范围上略有不同，如本试卷第3题，是一个积分题，尽管简单，但全国卷已经不考查了。
四、本考试卷考点分析表（考点/知识点，难易程度、分值、解题方式、易错点、是否区分度题）。
