How were conic curves discovered? And how to prove it? 15

Discovered by the ancient Greeks.

It is to cut on the cone from different directions.

Because the straight line y x a 1 is parallel to the hyperbolic x 2 a y 2 asymptote line y x 2, laugh at the brigade.

The straight line touches the stool y x-1 and intersects the right branch of the hyperbola at a point (the relationship with the left branch is separate). Big search.

That is to say: the straight line y x 1 and the hyperbola x 2 2 y 2 2 1 are intersecting, but there is only one intersection point!

y=x-1 is substituted for x 2-y 2=1, and x -(x-1) is noisy=2

i.e. 2x-1=2, x=3 2, y=1 2

Therefore, there is only one point of change in the sentence of lack of suspicion.

1. When the plane is parallel to the bus of the secondary cone, and it is not 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 vertex of the cone, the result is hyperbola.

6. When both sides of the plane and the secondary cone intersect and pass the vertex of the cone, 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.

Historical Background of Conic Curves:1. The mathematics of conic curves, some curves obtained by cutting cones in geometry.

2. Apollonius once called the ellipse a deficit curve, the hyperbola a hypercurve, and the parabola a homogeneous curve. In fact, Apollonius' use of pure geometry methods in his writings has achieved all the properties and results of conic curves in high school mathematics today.

Proof: Let pof=x, then.

tan∠pof

b afp po, it is easy to know the cheats, fp=b, po=a, through p to the x-axis perpendicular line, the vertical foot is q, not rotten flush is difficult to prove.

rt oqp rt opf,oq:op=op:of,oq=a 2 c, i.e. p on x=a 2 c, prove!

2. First of all, the eccentricity of the hyperbola is e>1, and the hyperbola has an intersection point with the left and right branches, according to the graph, that is, the relationship between the asymptote of the two hunger zones and FP.

Pass. 1. The inclination angle of the asymptote of the three quadrants must be greater than 45°, that is, b>a, a 22, e> 2, that is, the value range of e is. Thank you!

This student, this question is relatively simple. There are two ways to solve the problem, from the problem we know that the straight line equation of the right focus f(1,0) is y=2(x-1), and the simultaneous ellipse and the straight line equation get a(0,-2)b(5 3,4 3),|ab|You know, and then use the point to the straight-line distance formula to get d=2 5, S oab will come out You can also use two triangles OAF and obf, the bottom is the same as 1, and the height is added to |ya|+|yb|, which will be used in analytic geometry problems, can be summarized. Hope it helps.

tan＝pd/od

od*tan=pd

Pd is the height of the triangle to the land, and the required area has to be calculated by the kernel wisdom to find the Pd and then the area Bibi formula.

The definition of tan is the value of the right angle side of a right triangle compared to the other right angle edge in a right triangle.