High school math ellipse, ask a master or teacher to enter, if you answer well, I will add more poin

This problem is actually very simple, and the methods upstairs are too complicated, and this problem should be defined as an ellipse: that is, the sum of the distances from the points on the ellipse to the two focal points is 2a. Draw a picture.

Make a symmetry point f(-3,0) with respect to the y-axis

From the equation of the ellipse, we can see that a and f are the focal points of the ellipse and m is on the ellipse.

ma|+|mf|=2a=10

ma|=10-|mf|

ma|+|mb|=10-|mf|+|mb|=10-(|mf|-|mb|)

Again, -|fb|≤|mf|-|mb|≤|fb|i.e. -1 |mf|-|mb|≤1

9≤|ma|+|mb|≤11

ma|+|mb|The maximum and minimum values are 11 and 9 respectively, and I can guarantee that your answer is wrong.

Let the elliptic equation be: x a + y b =1 (a>b>0, because) e = 3 2, i.e., c a = 3 2, (a -b) a = 3 4, a = 4b

The first case: p(0,3 2) on an ellipse.

Since the center of the ellipse is at the origin and the focus is on the x-axis, the point p(0,3 2) is on the ellipse.

So b = 3 2, b = 9 4, a = 9

The elliptic equation is: x 9 + y (9 4) = 1 In the second case: p(0,3 2) is not on the ellipse (Note:

The solved b should be less than 3 2) x a + y b = 1, i.e. x 4b + y b = 1, x +4y = 4b, x = 4b -4y

Let the coordinates of the farthest point on the ellipse from p be (x,y), then we have:

4b -3(y +y) + 9 4), 4b is a fixed value, -3 (y +y) is a quadratic function of the opening downward, apparently the maximum value is obtained at y = -1 2 as 7, and when y = -1 2, 4b -3 (y + y) + 9 4) = 7

The solution is: b = 1 (in accordance with b<3 2), a = 4b = 4 elliptic equation is: x 4 + y = 1

It is known that the points a(3,0),b(-2,0) are the points in the ellipse x 25+y 16=1, and m is the moving point on the ellipse.

ma|+|mb|maximum and minimum values.

Solution: This is a conditional extremum problem, which we solve using the Lagrange multiplier method.

Let z=|ma|+|mb|= [(x-3) +y ]+x+2) +y ], where (x,y) is the coordinates of the moving point m.

Now the extremum of the function z when the condition f(x,y)=x 25+y 16-1=0 is required, and the function is used for this:

f(x,y)=√[(x-3)²+y²]+x+2)²+y²]+x²/25+y²/16-1]

f/∂x=(x-3)/√[(x-3)²+y²]+x+2)/√[(x+2)²+y²]+2λx/25=0...1)

f/∂y=y/√[(x-3)²+y²]+y/√[[x+2)²+y²]+2λy/16=0...2)

x²/25+y²/16-1=0...3)

Three-way simultaneous solution x, y, andwhere (x, y) is the coordinates of the extreme point, substituting.

z=|ma|+|mb|= [(x-3) +y ]+x+2) +y ], which yields the extremum.

Question 1. Let the coordinates of the m point (x,y).

Then the p coordinate is vertical (2x,y).

Because the forest code is p on a circle, (2x) 2+(y) 2=1 is the m trajectory square which fiber is (it can be seen that it is an ellipse).

Question 2. Let ab=bc=2c

According to the cosine theorem cos b = (ab 2 + bc 2 -ac 2 ) 2*ab*bc), ac is solved

Then, AC+BC=2A (in this case, A is the algebraic formula containing C) eccentricity = C A can be solved.

Child, the ellipse is not difficult, think about it yourself. Relying on others is never your own.

Elliptical symmetry Angle f1a0=30° Right triangle c a=1 2

Set |bf2|=m, then |bf1|=2a-m, in the triangle bf1f2, |bf1|2=|bf2|2+|f1f2|2-2|bf2||f1f2|cos120° ?2a﹣m）2=m2+a2+am

m=3/5 a

af1b area s = |ba||f1f2|sin60°a=10，c=5，b=5 。

1. Look at the picture to know af2=af1=a, and you can know that the angle af2o=30°, of2=c, obvious, e=c a=1 2

2. The first floor is the right solution.

Can you understand the above, a general idea.

Let's make a sketch first, look closely, and you're good to go, as long as you know the nature of the ellipse, ok (lalala.

Solution: by the title: in the ellipse b = 2, c = 2 root number 2 a = 2 root number 3

The equation for an ellipse is x 12+y 4 1 and b(0, 2) on an ellipse.

Let m(0-2) and n(x1,y1) give the linear equation to y kx 2

Lianli, eliminate y:

1+3k²)x²-12kx=0

=144k -4(1+3k)=4(33k-1) 0 i.e.: k root number 33 33 or k root number 33 33 In addition, from x1 = 12k (1 3k ) mn midpoint coordinates are q (6k (1 3k), 12k (1 3k ) 4).

And the equation for the perpendicular line of Mn is y=-1 k+2

Substituting the Q point coordinates into it and solving it: k 1

The equation for a straight line is either y x 2 or y x 2

According to the requirements of the problem, you can find the ellipse as x 2 12+y 2 4=1b(0,-2) on the ellipse, that is, one point in m,n, as long as you can find another point, it means that a straight line exists.

The problem translates to finding the distance from a point on the ellipse to a is 4

Consider setting m(0,-2),n(m,n).

m^2/12+n^2/4=1

m 2 + (n-2) 2 = 16 (m, n has a range) to find n = 0, n = -2

Therefore, n are the two vertices on the left and right, and we can find that there are two straight lines.

bc=a=2, ac=b, ab=c, so b+c=2a=4 is ac+ab=4, that is, the sum of the distances from a to b and c is a fixed value.

So the trajectory of point A is an ellipse, and the equation is x 2 4 + y 2 3 = 1 but abc is to form a triangle and b>a>c

So point A can only be in the second or third quadrant.

So the trajectory of point a is x 2 4 + y 2 3 = 1, (-2

Because if you don't have an equation, it's a bit difficult to give you the right idea. Since MF is perpendicular to the x-axis, it is known that the diameters of the two equations must be equal, i.e., p=b 2 a

Because the sum of the distances from the point on the ellipse to the two foci is constant, it is always 2a

The distances from the midpoint of each side of the square to the two diagonal fixed points are x 2 and (5) x 2, respectively

Therefore 2a = (x 2) + (5) x 2).

This is a quadratic function about y, y=-1 2 has a maximum value, but the function has a defined domain, that is, y may not be able to get -1 2, so we should compare b with 1 2 to discuss, if b is greater than him, then y=-1 2 takes the maximum value is equal to 7, if b is less than 1 2, then the maximum value is taken when -b, equal to 7

First of all, it can be known from the question: fa+fc=2fb=2b square aThen the coordinates of the two points are set, and the equation of the ac line is set.

He's over the f-point....The simultaneous ellipse subtracts yAnd then according to the Vedic theorem. You can list the equations.

This is an important application that is not yet understood. It's so tiring to type with your phone!

Passing by....Hehe, I think to do a good job of guiding and practicing questions, summarize the question types (many question types are available for guidance and practice!) It's really helpful to use good guidance and practice. But isn't the liberal arts class just ignoring it?Hehe.