It is one of the basic problems of analytic geometry to study the geometric properties of curves according to their equations and draw curves correctly. The purpose of analytic geometry is to list equations according to the conditions of curves. If it is a means of analytic geometry, then the purpose of analytic geometry is to study properties and draw pictures according to the equations of curves.

Next, we study the geometric properties of the ellipse according to the standard equation of the ellipse.

(1) range

Guide students to get inequalities from the standard equation, that is, it shows that in the rectangle surrounded by straight lines of an ellipse (as shown in the figure), pay attention to the graphic explanation and point out that when drawing points, you can't take points outside the range.

(2) Symmetry

Let the students read the geometric properties of ellipses in the textbook first.

Question: Why "when changing, or changing, or changing, or both, the solution of the equation remains the same, and the graph is symmetrical about the axis, axis or origin"?

In fact, in the curve equation, if the equation is unchanged, then when a point is on the curve, the symmetry point of the point about the axis is also on the curve, so the curve is axisymmetric. Similarly, the other two propositions can also be proved.

At the same time, it should be pointed out to students that if a curve has any two symmetries about axis symmetry, axis symmetry and origin, then it must have another symmetry.

Finally, it is emphasized that axis and axis are the symmetry axis of ellipse, and the origin is the symmetry center of ellipse, that is, the center of ellipse is the midpoint of focal line, and the symmetry axis is the focal line and its vertical line have nothing to do with coordinate system, so it is the inherent property of curve.

(3) Vertex

Guide students to analyze the intersection of ellipse and axis from the standard equation of ellipse, as long as it is clear that the point is the intersection of ellipse and axis; Suppose a point is the intersection of an ellipse and an axis. It should be emphasized that an ellipse has four vertices,,,.

At the same time, it should be noted that:

(1) The major axis and minor axis of the line segment and ellipse are equal to sum respectively;

The geometric meaning of (2) is the length of the long semi-axis of the ellipse and the short semi-axis of the ellipse.

(3) The vertex of ellipse is the intersection of ellipse and symmetry axis, and the vertex of general conic is the intersection of curve and its symmetry axis.

At this time, the teacher can make the following summary: from the scope, symmetry and vertex of the ellipse, draw more points. If you draw fewer points, you can get a more correct figure.

(4) Centrifugal rate

Because the concept of eccentricity is abstract, the teacher can directly give the definition of eccentricity:

The ratio of the focal length of an ellipse to the length of its major axis is called the eccentricity of the ellipse.

Firstly, the range of eccentricity is analyzed:

∵ , ∴ .

Combined with the chart, the influence of eccentricity on ellipse shape is analyzed:

(1) When it approaches 1, it approaches and becomes smaller, so the ellipse becomes flatter:

(2) When it is close to 0, it is close to 0 and thus close, so the ellipse is closer to the circle.

2. The definition of literal language

The ratio of the distance between a moving point and a fixed point and an alignment in a plane is a constant greater than 1. The fixed point is the focus of hyperbola, the fixed line is the directrix of hyperbola, and the constant e is the eccentricity of hyperbola.

2. Set the language definition

Let there be a fixed point M and a fixed point F on the hyperbola, and the distance from the point M to the fixed line is d, then the point set represented by the set {m |||| MF |/d = e, e > 1} is a hyperbola. Note: Fixed point F should be outside the alignment and the ratio should be greater than 1.

3. Standard equation

Let fixed point M(x, y) and fixed point F(c, 0), and the distance from point M to fixed line L: x = a 2/c is d, then | MF |/d = e>；; 1. The hyperbolic standard equation is (x? /a？ )-(y？ /b？ ) = 1, where a >；; 0，b & gt0，c？ =a？ +b？ This is the hyperbolic standard equation, with the center at the origin and the focus on the X axis. The hyperbolic standard equation with the center at the origin and the focus on the Y axis is: (y? /a？ )-(x？ /b？ ) = 1. Similarly: where a >；; 0，b & gt0，c？ =a？ +b？ .

Edit the simple geometric properties of this hyperbola.

1, the value range of a point on the trajectory: x≥a, x≤-a (focusing on the X axis) or y≥a, y≤-a (focusing on the Y axis). 2. Symmetry: Symmetry about coordinate axis and origin. 3. Vertex: A(-a, 0), A'(a, 0). At the same time, AA' is called hyperbola and the real axes B(0, -b), B'(0, b) of ∣AA'│=2a. At the same time BB' is called the imaginary axis of hyperbola and bb' = 2b. 4. Asymptote: focus on X axis: y = (b/a) X. Focus on Y axis: y = (a/b) X. Conic curve ρ=ep/ 1-ecosθ when e >: 1 represents hyperbola. Where p is the distance from the focus to the directrix, and θ is the angle between the chord and the X axis, so that θ can be obtained from 1-ecosθ=0, that is, the inclination of the asymptote. θ=arccos( 1/e) makes θ=0, which leads to ρ = EP/ 1-E, and X = ρ COS θ = EP/ 1-E makes θ=PI, which leads to ρ = EP/1+e. Find the abscissa of their midpoint (the abscissa of hyperbola center) X = (EP/1-e)+(-EP/1+e)/2 (pay attention to simplification) straight line ρ cos θ = (EP/1-e)+(-EP/65438). Let the rotated angle be θ′, then θ′ = θ-pi/2-arccos (1/e), then θ = θ′+pi/2-arccos (1/e) is brought into the above formula: ρ cos {θ′+pi/2-arccos (. The second definition: the ratio of the distance from point P to fixed point F on hyperbola to the distance d from point P to fixed line (corresponding directrix) is equal to the eccentricity of hyperbola. Point e. D(│PF │)/ line d (distance from point p to fixed line (corresponding to directrix)) =e 6, hyperbola focal radius formula (distance from any point P(x)Y on conic to focus), right focal radius: r=│ex-a│ left focal radius: r=│ex+a│ 7. Equilateral hyperbola-The length of the real axis and imaginary axis of hyperbola is equal, that is, 2a=2b and e=√2. At this time, asymptote equation is: y = x (regardless of whether the focus is on the x axis or the y axis) 8. * * Geometric expression of yoke hyperbola S': S: (x 2/a 2)-(y 2/b 2) = 1 s': (y 2/b 2)-(x 2/a 2) =1s'. C focus on the y axis: y = a 2/c10, path length: (in conic curve (except circle), chord out of focus and perpendicular to the axis) d = 2b 2/a655 chord length formula out of focus: d = 2pe/( 1-e 2cos 2θ). θ is the angle between the chord and the X axis] 12, and the chord length formula is d = √ (1+k2) | x1-x2 | = √ (1+k2) (x1-x2) 2 = + (y 1 - y2)？ ] After a little sorting, you can get: | ab | = | x1-x2 | √ (1+k? ) or | ab | = | y1-y2 | √ (1+1/k? )

Edit the standard formula and inverse proportional function of this hyperbola.

x^2/a^2-y^2/b^2 = 1(a & gt； 0, b>0) and the standard form of the inverse proportional function is xy = c (c ≠ 0), but the inverse proportional function is really obtained by rotating a hyperbolic function, because the symmetry axis of xy = c is y=x, y=-x, x 2/a 2-y 2/b 2 =/kloc. The y axis should be rotated by 45 degrees, and the rotation angle is a (a≠0, clockwise) (a is the inclination of hyperbolic asymptote), then X = xcosa+ysina Y =-xsina+ycosa takes a = π/4, then x2-y2 = (xcos (π/4)+ysin (π/4). 2 =(√2/2x+√2/2y)2-(√2/2y)2 = 4(√2/2x)(√2/2y)= 2xy。 And xy = cso x 2/(2c)-y.0) y 2/(-2c)-x 2/(-2c) =1(c < 0) It is proved that the inverse proportional function is actually a hyperbolic function, which is just another form of putting hyperbola in a plane rectangular coordinate system.

Edit this paragraph, hyperbola focus triangle area formula

If ∠F 1PF2=θ, then S△F 1PF2=b? Cot (θ/2) Example: It is known that F 1 and F2 are hyperbolas C: X? -Really? The left and right focus of = 1, point p is on c, ∠ F 1pf2 = 60, how small is the distance from p to the x axis? Use the hyperbolic focal triangle area formula to solve: S△F 1PF2=b? Cot (θ/2) = 1× Cot30, let the distance from P to X axis be H, then S△F 1PF2=? ×F 1F2×h=？ 2√2×h=√3，h=√6/2