Bai giang Elip

Chia sẻ bởi Trần Văn Bích | Ngày 02/05/2019 | 58

Chia sẻ tài liệu: Bai giang Elip thuộc Bài giảng khác

Nội dung tài liệu:

Ellipse
Subject: Mathematics

Short Description:
- Give the definiton of ellipse.
- Give the normal equation of ellipse and students find the way to prove the normal equation of ellipse in the text book.
- Problem 1,2,4 help students to know "How to build different ellipses ?”
- Problem 3 helps students to expand ellipse equation.
Objectives of Lesson: At the end of lesson, students can get to know the definition of elipse and the normal equation of ellipse.
Methods of teaching and activities: Simulation and use of ICT
How ICT is used : Simulation and tutorial
Objectives of ICT use : use of simulation to visualize a concept
Classroom management: Devide students into groups
Step-by-step:
Ellipse
Nguyen Van Hien
Le Qui Don High School, Quang Tri, Vietnam
Constructing an Ellipse
The locus of these two points is an ellipse.
Construct the two intersection points of the circles.
Construct another circle with center F2 and radius CB.
Construct a circle with center F1 and radius AC.
Given segments AB and two point F1,F2 . The moving point C is on the segments AB
Definition
Given two fixed points F1,F2 in the plane so that F1F2 = 2c >0 and a positive number a > c .
The locus of point M in the plane so that: MF1 + MF2 = 2a called an Ellipse.
Points F1, F2 called the focuses.
F1F2 = 2c is called focal length.
When M is on the Ellipse, then MF1 and MF2 are called focal radius of M.
Normal equation
Let us choose the square coordinate systems Oxy so that: F1(-c,0) , F2( c, 0).

(1) with b2 = a2-c2.
Ellipse (E) cut x-axis at D(-a,0) and E(a,0), DE is called the major axis . The major axis then is of length 2a
Ellipse (E) cut y-axis at P(0,b) and Q(0,-b),PQ is called the minor of the Ellipse . The minor axis is of length 2b
Equation (1) is called the normal equation of Ellipse.
The equation of the ellipse then can be written as
Note:
Problem 1
Given a point A and a circle (B; R) with the centre B and radius R so that 0 < AB The moving point C is on the circle. M is the intersection point of the segment BC and the perpendicular bisector of AC.
Find the locus of the point M.
Hint:
M is on the segment BC , MA = MC
The locus is the Ellipse with the focuses A, B and sum of focal radius is R
Hence it follows that the locus of C
is the ellipse (E) :
Problem 2
Given a circle with the diameter AB = 2R. The moving point M is on the circle. N is the foot of perpendicular line through M to AB. Point C on MN so that: NC = kMN (0 Find the locus of C.
Problem 3
Find the equation of the Ellipse (E) passing through point A(1,-1) with the focuses F1(-3,1) and F2 (1,1).
The equation of the Ellipse (E) is .
Hint:
B(-1,1) is the midpoint of F1F2 and F1F2 // Ox
Then The locus is the Ellipse with the
equation is


Problem 4
In the plane Oxy, given two concentric circles (C) and (C’) with centre O and the radii are a, b (a > b).
A moving ray Ot cutting (C) and (C’) at M and N respectively.
From M construct a parallel line to the Ox axis. From N construct a parallel line to the Oy axis.
Denote that the intersection of two constructed lines is E.
Find the locus of E.
Assessment
Given a point A and a circle (B;R) with the centre B and radius R so that 0 < AB < R. Find the locus of the centre of the circles passing through the point A and tangent to the circle (B)
b) Given A(4, 0) , B( - 4, 0) and R = 10. Find the equation of the locus
Solution:
a) M is the centre of the Circle (M) ,which passes through point A and contact with the circle (B) at the point N.
Then M lie on the segment BN and we have BN = R , MA = MN
MA+MB = MN+MB = NB = R = constant. Its follows that:
The locus is the Ellipse with the focuses A, B and sum of focal radius is R
b) If A (4, 0), B (- 4, 0) and R = 10

Equation of the locus is
Homework
The ladder is leaning against a brick wall and has a bucket resting on one of its rungs. Slowly, the ladder starts to slide down the wall. What path does the bucket trace?
3. Discuss and draw the graph of
16x2+9y2-32x+36y-92=0
2. Give a circle (A) in the circle (B) Find the locus of the centre of the circles contact with the circles (A) and (B).
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