Ask about sophomore math questions thank you, thank you .

First of all, the denominator should be written in the ...... after the score line

x^2/4+y^2/16=1

Hence a(0,4).

Let the midpoint be (x,y).

Since a is known, the coordinates of b are (2x,2y-4) and the coordinates of b satisfy the elliptic equation.

x^2+(y-2)^2/4=1

This is the trajectory equation for the midpoint, where the trajectory is an ellipse.

Can we express the coordinates of the midpoint with a parametric equation, and have we learned the parametric equation for the (cosa, 2sina+2) ellipse? Actually, it's easy to test. If you substitute cosa with 2sina+2, the final result on the left is (sina) 2+(cosa) 2=1

After the parametric equation represents b, the distance of ab is described.

d^2=(cosa)^2+(2sina-2)^23(sina)^2-8sina+5

Think of this equation as a quadratic function, sina is a variable, that is, find the maximum value of this function its opening is upward, and the axis of symmetry is x=4 3

Therefore, the maximum value is taken when sina = -1.

At this time b(0,0).

The maximum value is the distance from (0,4) to (0,0), which is 4

Let f0=(c,0),f1=(0,d),f2=(0,-d), where b 2=c 2 + d 2,d>0

So, in the equilateral triangle f0f1f2, c = 3 2, and d = 1 2The solution is b 2 = 4 3c 2 = 1, a 2 = 7 4 so the answer is.

4x 2 7) + (y 2 1) = 1 (x> = 0) with semi-ellipse (y 2 1) + (4x 2 3) = 1 (x< = 0).

By a 2 = b 2 + c 2, a>b, f0(sqrt(a 2-b 2),0), f1 (sqrt(b 2-c 2),0), f2(-sqrt(b 2-c 2),0), the triangle f0f1f2 has a side length of 1, so sqrt(3)*sqrt(b 2-c 2) = sqrt(a 2-b 2), so 3*b 2 = 4*c 2

And because sqrt(b 2-c 2)=1 2, so b=1 so c=sqrt(3) 2,a=sqrt(7) 2

Give it some points, it's hard work.

Question 1 (1) Since the images of y=f(x) and y=2-(1 x) are symmetrical with respect to the straight line y=x, we can know that the symmetry point of y=2-(1 x) is (0,2), and the symmetry point of f(x) is (2,0). i.e. f(x)=-1 (x-2). The sequence satisfies a(n+1)=f(a(n))(n n).

A(1) = 3 is obtained

a(2)=-1

a(3)=1/3

a(4)=3/5

and a(n+1)-a(n)=-1 (a(n)-2)-a(n)=-a(n)-1) 2 (a(n)-2).

As long as a(n)-2<0, there is a(n+1)>a(n) due to 0a(n).

h=32) I don't know what M is used for. Anyway, the first question also explained, when n 3, all numbers are dismantled in the interval (0, 1), so n = 3

Question 2: The original formula can be reduced to a(x 2+y 2)-bxy=1, which can be obtained from the mean inequality xy (x 2+y 2) 2.

bxy≤b(x^2+y^2)/2

i.e. (a-b 2) (x 2 + y 2) 1

then x 2 + y 2 1 (a-b 2).

1. The more conventional method is to set the equation and then solve it by substitution, but it is very troublesome.

Observe for yourself: the center of the circle b:x2+6x+y2-55=0 is (-3,0) and the inscribed circle passes the point (3,0).

What is the law of incision of two circles? That is, the centroid plus the radius of the minor circle equals the radius of the great circle! That is, the sum of the distance from the center of the small circle to (-3,0) and (3,0,) is the radius of the great circle 8, isn't this the definition of an ellipse, a=4,c=3,b=5 (1 2) It is easy to write this equation x 2 16 + y 2 5 = 1

I don't know if I can understand it, and if I can't understand it, then m me!

The tangent point is set to p in the center m of the moving circle m, which crosses the point a(3,0) and is inscribed with the circle b:x2+6x+y2-55=0

pm=papm=r-pb=pa

So PA+PB=R=8

So it's an ellipse.

c=3a=8/2=4

x^2/16+y^2/7=1

The deformation of the equation is: x 2-x + ab + b = 0

Because there is only one element.

So there is only one solution to the equation, and x=a

So discriminant = 0

Substituting x=a into the equation and calculating the discriminant.

1-4（ab+b）=0

a^2+ab+b-a=0

Because: 4(ab+b)=1, so ab+b=1 4, so: a 2+1 4-a=0 i.e. (a-1 2) 2=0 solution: a=1 2

So: b=1 6

There is only one solution, and there is a discriminant formula to obtain 1-4*(ab+b)=0 a is the solution of the original formula, and the |a^2+ab+b=a a=-1/2 b=1/2

Make sure you didn't make a mistake in the question: minus - 3sinx ?

Divide by cosx, cosx+2sinx) (2cosx - 3sinx) = (1 + 2tanx) (2 - 3tanx) = (1 - 6) (2 + 9) = -5 11

sinx=-3cosx, which can be solved in the equation and results in -5 11

Divide the top and bottom by cosx... Without further ado, see for yourself.

That is, any x r, 2x 2+(a-1)+1 2>0 is constant.

So use <0 for it and solve the range of a.

The negation of the proposition is that any x belongs to r, so that 2x 2+(a-1)x+1 2>0 is constant.

(a-1) 2-4<0, find the range of a.

1

That is, 2x 2+(a-1)+1 2>0 must have a solution.

Then beg for yourself.

Solution: If all the points (s,f(t))(s,t d) form a square, then the length of x of the defined domain and the length of the range are equal.

Define the length of x of the domain = |x1-x2|=√[(x1+x2)^2-4x1x2]

[(b/a)^2-4c/a]

[(b^2-4ac)/a^2]

The length of the range is from 0 to the maximum and is [-b 2 (4a)+c] [b 2 (4a)+c]= [(b 2-4ac) a 2]-b 2 (4a)+c=(b 2-4ac) a 2-ab 2+4a c=4b 2-16ac

4+a)b^2-4ac(4+a)=0

4+a)(b^2-4ac)=0

So, a+4=0, a=-4

The straight answer acre angle is aob, and the distance from the origin o(0,0) to the straight line 2ax+by=1 is 1 (2a 2+b 2)= 2 2, so there is the formula 2a 2+b 2=2, so that b 2<=2,- 2<=b<= 2;

The distance between p(a,b) and (0,1) is (a 2+(b-1) 2)= 1 2)b 2-2b+2), which can be seen from the range of the value of b