Mathematical conic curve problem solving, mathematical conic curve problem

It is known that F1, F2 is the left and right focus of the ellipse x 2 a 2 + y 2 b 2 = 1 (a>b>0), a is the point on the ellipse that is located in the first quadrant, af2 vector times f1f2 vector = 0If the eccentricity of the ellipse is equal to 2 2

1) Find the equation for the straight line ao (o is the coordinate origin).

2) The straight line ao intersects the ellipse at the point b, and if the area of the triangle abf2 is equal to 4 2, find the equation of the ellipse.

Solution: Consists of x a + y b = 0 (a>b > 0) e = 2 2

a²=2b²

x²/(2b²)+y²/b²=0

Vector af2 * vector f1f2 = 0

a is (b, b 2).

If the ellipse of the extended line ao intersects at the point b, then there is a vector oa + vector ob = vector ob is (-b, -b 2).

Straight line ab: y = 2x 2

Also. The equation for the straight line ao (o is the coordinate origin) is:

y=√2x/2

ao:tan aof2 =af2 of2=b 2 ac=1 root2:;

An elliptic equation with a linear equation in conjunction with a linear equation to find two solutions can be expressed by a or b A coordinates can be represented by a; That's pretty much the end of it.

Summary. Yes, only one question can be answered.

Yes, only one question can be answered.

For example, if the ellipse is a quarter of x square plus a third of y square, the focus is 1, 0, a straight line y through the right focus is equal to kx plus m to cross the ellipse in a and b, the known chord length formula is shown in the figure, the value of the δ is shown in the figure, and it can be known that m is equal to negative k, and the formula is substituted into the algebraic formula of the chord length about k, obviously it is positively correlated with k, and k has no size limit, and the chord length is obviously a maximum, so there is a problem with this Kirihe derivation? How to solve the problem?

There's nothing wrong with the chord length formula either.

The slope is infinite, but |x1-x2|However, it decreases as the slope increases.

And |x1-x2|It is obtained by simultaneous equations.

This calculation of yours is incorrect.

And this is determined by k and m, and is not proportional to k.

What if it's not (1, 0) but (2, 0)? The result is different.

It can only be said that the chord length changes with the magnitude of the slope k value after passing (1,0).

And from the equation of the straight line, we can see that the larger k is, the larger m is.

x1-x2|It is derived from the simultaneous equation, and does not depend on how to explain this special situation, after all, he conforms to this formula in this case, and the string length of the sock is indeed infinity when k is taken to infinity, and there must be a problem with the derivation, and the result of the derivation in this case should be in line with this situation.

I told you when K takes infinity|x1-x2|Take infinitesimal.

There's nothing wrong with the derivation, and there's nothing wrong with the formula, it's just that you're not deriving it correctly.

Variable analysis, incomplete consideration.

Solution: Eccentricity = root number 6 to 3

c 2 Lee, take this a 2=6 9=2 3

3c^2=2a^2

3(a^2-b^2)=2a^2

a^2=3b^2

The equation for the straight line over ab is: x a-y b=1

i.e.: bx-ay-ab=0

The origin is 3 to 2 from the root number, and there is: |-ab|/√b^2+a^2)=√3/2

4a^2b^2=3(a^2+b^2)

Substituting Min A 2 = 3b 2 into the above equation to solve the solution:

b^2=1a^2=3

The elliptic equation is: x 2 3 + y 2 = 1

The equation y is positive = 2px, and the equation is y =4x by substituting (1,2)4=2p

k(pa)=(y1-2)/(x1-1)=(y1-2)/(y1^2/4-1)=4(y1-2)/(y1^2-4)=4/(y1+2)

k(pb)=(y2-2) (x2-1)=(y2-2) repentance(y2 2 4-1)=4(y2-2) (y2 2-4)=4 (y2+2).

When the slopes of Pa and Pb exist and the inclination angles are complementary, K(Pa)+K(Pb)=04 (Y2+2)+4 (Y1+2)=0

4(y1+2+y2+2)/(y1+2)(y2+2)=0y1+y2=-4

kab=(y2-y1)/(x2-x1)=(y2-y1)/(y2^1/4-y1^2/4)=4(y2-y1)/(y2^2-y1^2)=4/(y1+y2)=-1

The square of a + 16 b(a-b) is greater than or equal to the square of a + 16*4 the square of a is greater than or equal to 2 times the root number 64=16 The condition for the equal sign to be established is b=a-b; The square of a is equal to the square of 64 a. Therefore e = the root number of 2 points 3

If you think about it this way, in the circular equation x has a defined domain, and x belongs to (0,a), so the equation should satisfy that there is a solution to (0,a), and then find it according to the distribution of real roots.

I tried e=1 2.,It turns out that he has a solution is 6.,I didn't envy him to figure out why the brother sent to 6.。。。

The radius is not that big

Let b(x0,y0), c(x,y) be rented by ca = know the hand.

15-x,-y) =,y-y0)

So. x0 = 5x 3-10 , y0 = 5y 3 1 and b on the circle x 2 +y 2 ==90000.

Substituting 1 into the above equation is the coarse equation of c.

Divide by 2 on both sides

1/2)/a²+(1/2)/b²=1

That is, x = 1 2 and y = 1 2.

So the fixed point is ( 2 2, 2 2).

The sum of the distances from the points (1,3 2) to the points f1 and f2 is equal to 4, so 2a = 4, and a = 2

So the equation for c becomes x 2 4 y 2 b 2 = 1 and brings the coordinates of the point (1, 3, 2) into the equation.

So the equation for the ellipse c is x 2 4 y 2 3 = 1 and the focal length is 4-3 = 1

So the focal coordinates are (-1 2,0) and (1 2,0).

Analytic geometry is a problem, and no one rushes to do it.

The coordinates of point m are easy to find (1,0).

Simultaneous y=- x( 5) 2+( 5) 2 and elliptic equation (x a) 2+(y b) 2=1 can be obtained.

x 2)((1 a) 2+5(1 b) 2 4)-5x (2(b 2))+5 (4(b 2))-1=0 The solution of this equation x1,x2

5/(2(b^2))+25/(4b^4)-4((1/a)^2+5(1/b)^2/4))(5/(4(b^2))-1))^

5/(2(b^2))+4/a^2+5/b^2-5/(ab)^2)^

Vector am=2 vector mb means.

x1-1)/(1-x2)=2

x1-1=x-2x2

x1+2x2=3

Substituting the expressions of x1 and x2 into the above equation yields it.

6/a^2=(4/a^2+5/b^2-5/(ab)^2)^

36/a^4=4/a^2+5/b^2-5/(ab)^2=4/a^2+5/b^2(1-1/a^2)

B 2 = 5a 2 (a 2-1) 4 (9-a 2).

Obviously, the condition of b 2>0 is satisfied, so 1 takes into account ba 2<41 9

Therefore, the value range of a is 1