Typical example
Definition of ellipse
Example 1
A straight line passing through a focal point F 1 of the ellipse 4x2+Y2 = 1 intersects with the ellipse at points A and B, then the circumference of △ABF2 formed by A and B and another focal point F2 of the ellipse is
[ ]
Brief description:
∵|AF 1|+|AF2|=2，
|BF 1|+|BF2|=2，
∴|af 1|+|bf 1|+|af2|+|bf2|=4，
That is | AB |+| AF2 |+| BF2 | = 4.
Choose B.
Comments:
This problem is to find the perimeter, but it is actually defined by an ellipse.
Example 2
The point m is a point on the ellipse, and the two focuses of the ellipse are F 1, F2. And 2a= 10, 2c=6, and point I is △MF 1F2.
Solution:
As shown in the figure, I is the heart of △MF 1F2,
∴∠ 1=∠2,
Comparing ① and ②, applying the equal ratio theorem, we can get
Comments:
The three steps of this problem all use the definition of ellipse, the theorem of internal bisector and the theorem of equal ratio. The equal ratio theorem is a bridge between the line segment ratio of the inner bisector and the first definition of ellipse.
Example 3
It is known that the two focuses of an ellipse are F 1, F2, and the point m is a point on the ellipse (not on the straight line F 1F2), ∠ F 1F2 = θ, |F 1F2|=2c, | MF/kloc-0.
Solution:
Draw a conclusion from cosine theorem
(2c)2=|F 1F2|2
= | MF 1 | 2+| MF2 | 2-2 | MF 1 | | MF2 | cos∠f 1mf 2
=(| MF 1 |+| MF2 |)2-2 | MF 1 | | MF2 |( 1+cosθ)
=(2a)2-2 | MF 1 | | MF2 |( 1+cosθ)
Comments:
Example 4
It is known that Equation 2 (K2-2) x2+K2Y2+K2-k-6 = 0 represents an ellipse, and the value range of realistic number k is obtained.
Solution:
According to the meaning of the question, get
Comments:
To solve this kind of problem, we should pay attention to two types of ellipses and the difference between ellipses and circles.
Example 5
Solution:
Let the elliptic equation be ax2+by2 = k,
①
Comments:
We don't know the type of ellipse in this problem, so we use this "fuzzy" method to simplify the calculation.
Example 6
Analysis:
Solution:
Let |PF 1|=m, |PF2|=n, m+n = 20,
That is m2+N2-Mn = 144.
( 1)
∴(m+n)2-3mn= 144.
Comments:
The above method uses the definition of ellipse and cosine theorem, and is often used to solve the triangle problem in ellipse.
Find the maximum value of | pf 1 || pf2 |.
Solution:
∫a = 10，
∴|PF 1|+|PF2|=20.
"=" is true if and only if |PF 1|=|PF2|.
∴| PF 1 || PF2 | The maximum value is 100.
Example 7
certificate
(1)∵P0 is outside the ellipse,
(2)P0 is within an ellipse,
Comments:
1. The knowledge points involved in this question are elliptic equations and coordinate concepts.
2. This is a common knowledge point, and it is not difficult to prove it as long as you understand the concepts of coordinates and curve equations.
Example 8
, find the minimum value of | am |+2 | MF |, and find the coordinates of m at this time.
Analysis:
According to conventional thinking, let M(x, y), then
M is on an ellipse, and Y can be represented by X, so that | am |+2 | MF | can be represented as a univariate function of X, and then the minimum value of this function can be found. Although this method seems feasible, it is very difficult to operate in practice, but we can convert the second definition of ellipse into the distance from point to straight line, as shown in the figure.
∴|AM|+2|MF|=|AM|+d
Since point A is in the ellipse, if a is AK⊥l and k is the vertical foot, it is easy to prove that |AK| is the minimum value of | AM |+D, and its value is 8-(-2) = 10.
Example 9
[ ]
A. ellipse
B. hyperbola
C. line segment
D. parabola
Brief description:
That is, the ratio of the distance from point P(x, y) to fixed point F( 1, 1) to the distance from fixed line L: x+y+2 = 0.
The trajectory of point P is an ellipse, so choose A.
Comments:
This question is wonderful: the beauty lies in the use of the second definition of ellipse, which can not be used directly, and the answer will not be known until it is deformed. It will be very troublesome to solve the problem with a double-sided square.
Example 10
Set out
[ ]
A.8
Brief description:
draw
|PF 1|+|PF2|=2a= 10，
∴|PF 1|=2.
∴|pf2|= 10-|pf 1|= 10-2=8.
Choose a.
Comments:
This problem is a comprehensive application of the first definition and the second definition of ellipse.
Example 1 1
As shown in the figure, the center of the ellipse is O, F is the focus, A is the vertex, the directrix L intersects the OA extension line at B, P and Q are on the ellipse, PD⊥l is at D, and QF⊥OA is at F, then the eccentricity of the ellipse is
[ ]
A: 0
B.2
C.2
D.5
Answer:
D.
Comments:
This topic uses eccentricity flexibly to deepen the understanding of the second definition of ellipse.
Example 12
Then | pf 1 | = a+ex0, | pf2 | = a-ex0.
Prove:
Defined by the second ellipse, we must
Comments:
In some books, the above conclusion is called focal radius formula. It is unscientific to do it according to the requirements of the textbook of People's Education Press, and it is easy to fall into a simple memory formula, ignoring the understanding and application of the second definition of ellipse. Because it is convenient to describe, the focal radius formula is still used later, but attention should be paid to understanding.
In fact, the above conclusion is an extension of the second definition of ellipse. Mastering the second definition of ellipse and the positional relationship between a point and a straight line is easy to deduce and remember. When used, you can add the prefix "according to the second definition of ellipse" to apply.
|PF 1|=a+ex0，|PF2|=a-ex0，
Example 13
Analysis:
Just solve the equation.
This method is natural, but it needs a lot of calculation, so we need to find a new solution from another angle.
Solution:
Defined by the second ellipse, we must
Comments:
If you fully understand the second definition of ellipse, you can remember the relevant conclusions.
Geometric properties of ellipse
Example 1
It is known that the center of the ellipse is at the coordinate origin, the focus is on the same coordinate axis, the eccentricity e=0.6, and the ellipse passes through point A (5,4). Find the elliptic equation.
Solution:
Comments:
Pay attention to two kinds of elliptic equations.
Example 2
Given the elliptic equation (1-m) x2-My2 = 1, find the length of the major axis.
Solution:
Elliptic equations are the second type.
Comments:
The type of equation is very important.
Example 3
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A the ellipse area decreases and the distance between the focus and the corresponding directrix increases.
B, the ellipse area decreases, and the distance between the focus and the corresponding directrix decreases.
The distance between the focus and the corresponding directrix increases with the increase of ellipse area.
D, the distance between the focus and the corresponding directrix decreases with the increase of ellipse area.
Choose B.
Comments:
The shape of an ellipse is flat and round. How can we describe its flatness?
When E is closer to 1 (i.e. increasing), C is closer to A, so B is smaller, so the ellipse is flatter.
If a is constant, the smaller the ellipse area is, S =πab；; The smaller the distance between the focus and the corresponding directrix, the focus moves outward.
The closer e is to zero (that is, decreasing), the closer c is to zero, so the bigger b is, the closer the ellipse is to a circle.
If a is constant, the larger the ellipse area is, S =πab；; The greater the distance between the focal point and the corresponding directrix, the focal point moves inward.
Note: The change of E number above reflects the flatness of ellipse. If the two focal points coincide with the origin, that is, a=b, then when c=0, the qualitative change of the graph is no longer an ellipse, but a circle X2+Y2 = A2.
Example 4
Is the distance from m point to two focal points F 1 and F2 equal to the average? And explain why.
Solution:
Suppose there is a point on the ellipse that satisfies the problem meaning P(x0, y0).
L: x =-4，| Mn | = | x0+4 | = x0+4。
If |MN| is the equal ratio median of |MF 1| and |MF2|,
Defined by the second ellipse, we must
Because the point M(x0, y0) is on the ellipse, it should be -2 ≤ x0 ≤ 2, and obviously the abscissa is x0 =-4.
Comments:
The essence of solving this problem is to reduce to absurdity.
Example 5
The earth's orbit around the sun is an ellipse, and the sun is at one of its focal points. The distance from the perihelion to the sun is1440,000 km, and the distance from the apohelion to the sun is1490,000 km. Find the eccentricity and orbit equation of this orbit.
Solution:
Take the length of millions of kilometers as the unit, and establish the coordinate system as shown in the figure.
∴a-c= 144,a+c= 149，
∴a= 146.5，c=2.5。
Comments:
Let P(x0, y0) be any point on the ellipse, then | pf2 | = a-ex0,-a ≤ x0 ≤-a.
∴(|PF2|)max=a-e (-a)=a+c，(|PF2|)min=a-e a=a-c
Maximum problem
Example 1
Find the maximum area of an ellipse with the long axis as the base.
Solution:
Let the elliptic equation become
Comments:
1. The knowledge points involved in this question are: the equation of ellipse (parameter equation), the area formula of trapezoid, and the processing method of the maximum value of trigonometric function.
2. The solution of the maximum problem is generally to select design variables (suitable independent variables), establish an objective function (that is, a mathematical model), and then apply functional knowledge and inequalities to find the maximum value of the objective function. Here, how to build a mathematical model is the key, and skilled mathematical language is the basic tool for building a model. Function knowledge and inequality are the basis of finding the maximum value.
Example 2
Solution:
Pass b to BN⊥l of n and a to AM⊥l of m.
From the nature of the ellipse, we know that
Comments:
According to the flatness of the graph, point B can be determined to minimize U 。
Example 3
Ellipse, find out where the m point is, the longest axis of the ellipse is the shortest, and find out the equation of this ellipse.
Analysis:
The length of the major axis of an ellipse is the sum of the distances from a point on the ellipse to two focal points. In this way, finding the ellipse with the shortest long axis through the point m on the straight line L is transformed into finding the point on the straight line L, so that the sum of the distances from this point to the two focal points F 1 and F2 is minimized.
Solution 1:
a2= 16，b2= 12，
∴c2=a2-b2=4.
Solution 2:
The circle should be tangent to the straight line l, m is the tangent point,
Eliminate y to get (a2+B2) x2-8a2x+16a2-a2b2 = 0.
∴△=64a4-4(a2+b2)( 16a2-a2b2)=0.
Simplified A2+B2 = 16.
( 1)
∴a2-b2=4.
(2)
Simultaneous equations from ① and ②, a2= 10, B2 = 6.
Comments:
Among the ellipses that focus on F 1 and F2 and pass through a point on the straight line L, the longest axis of the ellipse tangent to the straight line L is the shortest, because all points on the straight line L are outside the ellipse except the tangent point, and the sum of the distances from a point outside the ellipse to the two focuses is greater than 2a. This conclusion can be proved by plane geometry knowledge, and it can also be proved that the sum of the distances from one point to two focuses on an ellipse is less than 2a.
Focus radius problem
Example 1
Solution:
exi)(i= 1、2、3)。
∴ The abscissa of three points becomes arithmetic progression, which is the necessary and sufficient condition that the focal radius of this point becomes arithmetic progression.
Comments:
1. The knowledge points involved in this question are the focal radius, arithmetic progression and necessary and sufficient conditions of an ellipse.
2. Only by mastering the focal radius of the ellipse can we successfully pass this kind of basic problem, which shows the importance of mastering the relevant elements while mastering the standard equation.
Example 2
Distance into arithmetic progression.
(1) found x1+x2;
(2) Prove that the perpendicular line of AC passes through a certain point, and find the coordinates of this fixed point.
Solution:
(1) From the semimajor axis a=5, semiminor axis b=3 and semifocal length c=4 of the elliptic equation, according to the unified definition of conic curve, we get
∵|AF|+|CF|=2|BF|，
Therefore, x 1+x2 = 8.
Comments:
1. The knowledge points involved in this question are: elliptic equation, unified definition of conic curve, midpoint formula, point skew, arithmetic progression and linear system equation.
2. According to the unified definition of conic curve, arithmetic progression equation containing x 1 and x2 is formed by |AF|, |BF| and |CF|, and the formula of focal radius of ellipse is deduced, which is an indispensable mathematical language. It is the key to prove that the AC midline crosses a certain point. Because A and C are moving points on an ellipse and the center line of AC is a straight line system, the equation of this straight line system can be solved.
Lines and ellipses
Example 1
(1) has two things in common;
(2) There is only one thing in common;
(3) There is nothing in common.
Solution:
∴△=( 12k)2-4×9×(6k2-8)=-72(k2-4).
(1) △ > 0 means -2 < k < 2, the straight line and the ellipse have two common points.
(2) When △ = 0, that is, k =-2 or k=2, a straight line and an ellipse have only one common point.
(3) when △ < 0, that is, k > 2 or K.
Comments:
△ method for judging the general step convergence of the positional relationship between a straight line and an ellipse;
(1) simultaneous equation.
(2) Elimination into quadratic equation.
③ Calculate delta = b2-4ac.
(i) When △ > 0, two common points of a straight line and an ellipse are said to intersect.
(ii) When △=0, the straight line and ellipse have one and only one common point. At this time, the straight line and ellipse are called tangency.
(iii) When △ < 0, there is no common point between the straight line and the ellipse. At this time, the straight line and the ellipse are separated.
note:
There are two ways to distinguish the positional relationship between a straight line and a circle: the △ method and the "D-R" method, in which the "D-R" method is the simplest and most commonly used method, and there is only one way to distinguish the positional relationship between a straight line and an ellipse, namely the △ method.
Example 2
Symmetry on a straight line y = 4x+m 。
Solution 1:
∴△=(-8b)2-4× 13×( 16b2-48)>0.
( 1)
Let the midpoint of PQ be M(x, y).
(2)
Substitute ② into ① to get.
Solution 2:
Let P(x 1, y2), Q(x2, y2) be two qualified points on ellipse C, and M(x, y) be the midpoint of PQ.
The two expressions are subtracted, and 3 (x1-x2) (x1+x2)+4 (y1-y2) (y1+y2) = 0.
∵x 1≠x2，x 1+x2=2x，y 1+y2=2y，
∴M(-m,-3m).
Point m should be in ellipse c, so
Comments:
To solve this problem, we must deeply understand the meaning of symmetry.
Example 3
The length of AB.
Solution:
The right focus f of the ellipse (1,0),
Let the straight lines L: y = x- 1, A(x 1, y 1), B(x2, y2), ∴ y 1-y2 = x 1-x2,
Comments:
(1) If the area of △F 1AB is further calculated for this problem, as shown in the figure, F 1 (- 1, 0).
∴ the distance from point F 1 (- 1, 0) to the straight line L: Y = X- 1
There are usually two methods to find the area of △F 1AB:
( 1)S△f 1AB = S△f 1F2A+S△f 1F2B，
(2) The straight line L in this question is the right focus of the ellipse, and the chord cut is "focus chord". For "focus chord", besides the method of finding chord length mentioned above, we can also consider the following research on | AB | = | AF2 |+| BF2 | to continue our research.
Example 4
It is known that the center of the ellipse is at the coordinate origin o, the focus is on the coordinate axis, and the straight line y = x+ 1 intersects the ellipse at
Analysis:
It is not clear whether the ellipse focus is on the X axis or the Y axis in the question setting. So when choosing the standard equation form, it is not clear whether A > B or A < B, which is one of them. On the other hand, the problem involves two conditions of intersection p and q, which simplifies the calculation.
Solution:
Let the equation of ellipse be
According to the meaning of the question, the coordinates of point P and point Q satisfy the equation.
Substitute ② into ① to get.
(a2+b2)x2+2a2x+a2( 1-b2)=0。
③
Let the two roots of Equation ③ be x 1 and x2, respectively, then the intersection of straight line Y = X+ 1 and ellipse P(x 1, x 1) and Q(x2, X2+ 1).
Pack up, take it
Solve X 1+X2 and x 1+x2 to obtain
Solve it and get it.
Therefore, the elliptic equation is
Comments:
This topic is a clever combination of ellipse and Vieta's theorem.
Trajectory and synthesis
Example 1
It is known that the point P moves on the straight line X = 2, the straight line L passes through the origin and the point A (1 0) perpendicular to the OP, and the straight line M intersects with the straight line L of the point P at the point Q, thus finding out the trajectory equation of the point Q and pointing out the name of the trajectory and its focal coordinates.
Solution 1:
Let Q(x, y), P(2, t)
∵OP⊥OQ，
∴ty=-2x.①
∵Q, a, p three-point * * * line,
That is y = t (x- 1).
If t≠0, then t is eliminated by ① and ②, so that.
2x2+y2-2x=0(y≠0)。 (※)
If t = 0, then p (2 2,0), l: x = 0.
※ ∴ q (0 0,0) also satisfies the formula.
To sum up, the trajectory equation of moving point Q is
Solution 2:
When l⊥x axis, then q (0 0,0).
When l is not perpendicular to the x axis, let l: y = kx, where k ≠ 0.
solve an equation
Let Q(x, y), then y = kx. ①.
Merge ① and ②, and eliminate K to get 2x2+y2-2x = 0 (y ≠ 0).
Q (0 0,0) also applies to the above formula.
∴ The trajectory equation of point Q is
The track name and focus coordinates are the same solution.
Comments:
The motion of point Q can be regarded as the motion of point P, so the ordinate of point P can be used as a parameter to describe the motion law of point Q; The movement of the point Q can also be regarded as the movement of the straight line L around the origin, so the inclination or slope of the straight line L can be selected to establish the relationship between the abscissa and ordinate of Q. 。
Example 2
The other point Q is on OP, which satisfies | OQ || OP | =| or | 2 (as shown in Figure 2). When point P moves along straight line L, find the trajectory equation of point Q, and explain what curve the trajectory is.
Analysis:
The given equation | OQ |||| OP | =| or | 2 involves not only the moving point Q we need, but also the moving points P and R. In order to get the trajectory equation of Q, the relationship between the coordinates of P, R and Q can be determined by the combination of straight lines of O, Q, P and R, the combination of R points and ellipses, and the combination of P points and straight lines. It is also a good method to describe the coordinates of q, r and p with XOP = θ. If q, p and r are projected on the x axis, the problem can be expressed as follows.
Solution 1:
Let's assume that point q is not at the origin. Let's assume P(xP, yP), R(xR, yR) and Q(x, y), where x and y are not both 0.
When point P is not on the Y axis, from the straight line of point R and O, Q and R on the ellipse, we can get
Solve it and get it.
From the straight line of point P and O, Q and P on the straight line L, we can get
When the point P is on the Y axis, P (0 0,8), R (0 0,4) and Q (0 0,2) are easy to verify that ①, ②, ③ and ④ are all true.
From the topic | oq ||| op | =| or | 2, it is concluded that.
Substitute ①, ②, ③ and ④ into the above formula, and you will get
∵x is the same as xP, and Y is the same as yP;
③ and ④ ∴ 2x+3y > 0.
Solution 2:
Let's assume that point q is not at the origin. Let's assume P(xP, yP), R(xR, yR) and Q(x, y), where x and y are not all zero.
Let ∠ XOP = α, there are the following kinds:
xP=|OP| cosα，yP = | OP | sinα
xR=|OR| cosα，yR = | OR | sinα
X=|OQ| Coase α, y=|OQ| Coase α.
Equation | oq ||| op | =| or | 2 is given by the above formula and problem.
From the point p on the straight line L and the point on the ellipse, we get
Substitute ①, ②, ③ and ④ into ⑤, ⑤.
Pack up, take it
An ellipse parallel to the X axis, with the origin O (0 0,0) removed.
Solution 3:
Q is not at the origin. Let P(xP, yP), R(xR, yR) and Q(x, y), where x and y are not zero at the same time.
∵| OQ | | OP | = | or |2
According to the meaning of the question, x, xP and xR have the same number, and y, yP and yR have the same number, so
Because the point P(xP, yP) is on the straight line L and O, P, Q, it is obtained that
Substituting ① and ② into formula (※), we get
Ellipse with axis parallel to X axis, remove the origin O (0 0,0).
Comments:
This topic is the finale of 1995 national college entrance examination mathematics test, and this year's liberal arts test is its sister topic:
Know that point Q is on OP and satisfies | OQ || OP | =| or | 2. When Q point moves on L, find the trajectory equation of Q point and explain what curve the trajectory is.
Example 3
Fold it into a straight dihedral angle along the X axis, and find the angle formed by the connecting line of AB and the X axis.
Solution:
B leads BC‖OX to intersect the ellipse in C. B and C are symmetrical about Y, and A and C are symmetrical about X..
∴| AD | = | DC | and both AD and DC are perpendicular to the X axis.
After the coordinate plane is folded, the positional relationship of points or lines in the same plane remains unchanged, and there is still | AD | = | DC | and both AD and DC are perpendicular to the X axis.
∠ADC is the plane angle of dihedral angle,
The AD⊥ plane BOC is shown in fig. 3,
∵BC⊥CD According to three mutually perpendicular theorems, the alternating current of ⊥ BC is obtained.
In Rt△ABC, before isomorphism |BC| is 2|OA|cosα,
Comments:
(1) The knowledge points involved in this topic are: elliptic equation, trigonometric function concept and the relationship between the angles of right triangle, the plane angle of dihedral angle, the theorem of three perpendicular lines and the angle formed by lines on different planes.
(2) This is a comprehensive problem of analytic geometry and solid geometry. Draw an intuitive view according to the topic to assist spatial imagination. After folding, the relative positions of points and straight lines in the same plane remain unchanged, while the relative positions of points and straight lines in different planes change. This should be paid attention to when judging the positional relationship.