Summary of conic curves, summary of knowledge points of conic curves

1. When the flat surface is parallel to the bus of the secondary cone, and is not more than the apex of the cone, the result is a parabola.

2. When the plane is parallel to the bus of the secondary cone plane and passes the vertex of the cone, the result degenerates into a straight line.

3. When the plane only intersects one side of the secondary cone, and does not exceed the vertex of the cone, the result is an ellipse.

4. When the plane only intersects one side of the secondary cone, and does not exceed the vertex of the cone, and is perpendicular to the axis of symmetry of the cone, the result is a circle.

5. When both sides of the plane and the secondary cone intersect, and the cone vertex is not exceeded, the result is hyperbolic (each of which is the intersection line between the conical surface and the plane of one of the secondary cones).

6. When both sides of the plane and the secondary cone intersect the old plexus, and pass the cone vertex, the result is two intersecting straight lines.

7. When the plane and the two sides of the secondary cone plane do not intersect, and the vertex of the conic is passed, the result is one point.

One is the entry point: from the beginning, in order. Reverse order from tail. From the middle, the top is connected to the bottom.

The second is the transformation method: spread method, set but not seek, chord length formula, discriminant, Vedic theorem, point-to-straight line distance formula, Pythagorean theorem, sine theorem, cosine theorem, area formula, focus triangle, definition, can not be played.

The distance from the point on the ellipse to the focal point is called the focal radius.

The distance to the left focal point is a+ex

A and E are both deterministic.

and -a<=x<=a

So x=-a

a+ex minimum=a-ea=a-c a*a=a-cx=a, maximum=a+c

The distance to the right focal point is a-ex

The same goes for the above conclusion.

There is a question of perigee and apogee in the book, so you should go and read it.

The slope pin of the straight line l is set to k

l:y=kx-2 substitution x 4+y =1

Get: x +4(kx-2) = 4

That is: (4k +1) x -16kx + 12 = 0 set a(x1,y1),b(x2,y2).

256k²-48(4k²+1)>0

64k²-48>0,4k²-3>0

According to Veda's theorem.

x1+x2=16k/(4k²+1),x1x2=12/(4k²+1)ab|= 1+k )*x1+x2) -4x1x2] (1+k )*256k (4k +1) -48 (4k +1)] 1+k )*64k -48) (4k +1) ]o point to the distance of the line l d=2 (1+k )

Vector on=oa+ob

The quadrilateral oanb is a parallelogram.

soanb=2sδaob

ab|*d2*√[64k²-48)/(4k²+1)²]8√[(4k²-3)/(4k²+1)²]

Let 4k -3=t>0

The root number is under the formula.

u=t/(t+4)²=t/(t²+8t+16)1/(t+16/t+8)

Deficit delay t+16 t 2 16=8

But only if t = 16 t, t = 4, i.e. k = 7 4 take the equal sign.

u≤1/(8+8)=1/16

The maximum value of SOANB is 2

At this point k = 7 2

The equation for a straight line which roll l is y= 7 2x-2

Well, mom, the method is third-party.

1) Let the coordinates of point A (x1, y1) and point b coordinates (x2, y2), the slope of the straight line where af1 is k, because it passes the point f1, so the linear line equation is y=k(x+1), y=k(x+1) ......1)

x^2)/2+y^2=1……(2)

The unary quadratic equation of y is solved (1 2k 2+1)*y 2-y k-1 2=0, because a is the point above the x-axis, taken.

The positive root of the equation dust let, y1=[1+( 2+2k 2)]*k (1+2k 2)

In the same way, since bf2 is parallel to af1, the slope of the straight line is k, and because it passes the point of f2, the equation for the straight line is.

y=k(x-1),y=k(x-1)……3)

x^2)/2+y^2=1……(4)

The unary quadratic equation of y is solved (1 2k 2+1)*y 2+y k-1 2=0, because b is the point above the x-axis, taken.

The positive root of the equation, y2=[-1+( 2+2k 2)]*k (1+2k 2)

y1=k(x1+1), y2=k(x2-1), the two formulas are added together, y1+y2=k(x1+x2)......5)

af1-bf2 = 6 2, af1=ex1+a, bf2=a-ex2, af1-bf2=e(x1+x2)=

6 2, x1+x2=( 6 2) e= 3, substitute the obtained y1, the positive y2 value and x1+x2 = 3 into (5), and the solution is k= 2 2 or -

Substituting the k value into y1 and y2, when the first branch regrets considering k= 2 2, y1=( 6+ 2) 4, y12=( 6- 2) 4, x1=(

3-1)/2,x2=(√3+1)/2；k=- 2 2, the y value is rounded off negatively.

2) Substituting the k value into y1 and y2, k = 2 2, y1 = ( 6 + 2) 4, y12 = ( 6- 2) 4, x1 = ( 3-

1) 2,x2=( 3+1) 2,kaf2=y1 (x1-1)=(2 6+3 2) (6),kbf1=y2 (x2+1)=(3 2+2 6) (6), af2 linear equation is y=kaf2(x-1), bf1 linear equation is y=kbf1(x+1), and the p coordinates of the intersection of two straight lines are (

3 2, 6 12), p, f1, f2 are fixed values, and pf1+pf2 are fixed values.

The answer isn't necessarily right, hope it helps.