S=π (pi) ×a×b (where A and B are the lengths of the major axis and minor axis of the ellipse respectively) or S=π (pi) ×A×B/4 (where A and B are the lengths of the major axis and minor axis of the ellipse respectively).

Perimeter formula of elliptic circle

Ellipse circumference has no formula, just an integer or infinite expansion. The exact calculation of ellipse circumference (L) requires the summation of integral or infinite series. Such as l = ∫ [0, π/2] 4a * sqrt (1-(e * cost)? )dt≈2π√((a？ +b？ ) /2)[ Approximate circumference of ellipse], where a is the long semi-axis of ellipse and e is eccentricity. Eccentricity of an ellipse is defined as the ratio of the distance from a point on the ellipse to a focal point to the distance from the point to the directrix corresponding to the focal point. Let the distance from the point P on the ellipse to a focus be PF and the distance from the corresponding directrix be PL, then e=PF/PL.

The directrix equation of ellipse

x= a^2/c

Eccentricity formula of ellipse

e = c/a(0 & lt； E< 1, because 2a >；; 2c. The greater the eccentricity, the flatter the ellipse; The smaller the eccentricity, the closer the ellipse is to the circle. Focal length of ellipse: the distance between the focal point of ellipse and its corresponding directrix (such as focal point (c, 0) and directrix X =+A 2/c) is b 2/c.

Elliptic focal radius formula

The focus is on the x axis: | pf1| = a+ex | pf2 | = a-ex (f1,f2 is the left and right focus respectively). The ellipse radius passing through the right focus r=a-ex passes through the left focus r=a+ex. The focus is on the y axis: | pf 1 | = a-.

Position relationship between point and ellipse

Point M(x0, y0) ellipse x 2/a 2+y 2/b 2 = 1 in the circle: x0 2/a 2+y0 2/b 2 1

The positional relationship between straight line and ellipse

Y = kx+m1x2/a2+y2/B2 =12x2/a2+(kx+m) 2/B2 =1tangency delta = 0 phase separation delta.

Slope formula of ellipse

X 2/a 2+y 2/b 2 =1On an ellipse, the tangent slope of point (x, y) is-(b 2) x/(a 2) y If the formula of the area of the ellipse focus triangle is ∠ f1pf 2,

Application of editing elliptic parameter equation in this paragraph

When solving the maximum distance from a point to a fixed point or a straight line on an ellipse, the problem can be transformed into a trigonometric function problem by using parametric coordinates, where x=a×cosβ, y=b×sinβ a is half the length of the long axis. Because the figure obtained by cutting a plane cone (or cylinder) may be ellipse, it belongs to a conical section. For example, if a cylinder is cut to get a cross section, which proves to be an ellipse (using the first definition above): two hemispheres with the same radius as the cylinder are squeezed from both ends of the cylinder to the middle, and stop when they touch the cross section, then two common points will be obtained, which are obviously the tangent points of the cross section and the sphere. Let two points be F 1, F2. For any point P on the cross section, the generatrix Q 1 and Q2 passing through P are cylinders, and the great circles tangent to the sphere and cylinder intersect with Q 1 and Q2 respectively, so PF 1=PQ 1, PF2=PQ2, so PF 1. Similarly, centering on F 1 and F2, it can also be proved that the oblique section of the cone (which does not pass through the bottom) is an ellipse: ellipse c: x 2/a 2+y 2/b 2 =1(a > b > 0) eccentricity is √6/3, and the distance from one end of the short axis to the right focus is √ 3. 1. Find the equation of ellipse C2. The straight line L: Y = x+ 1 intersects with the ellipse at two points A and B, and P is a point on the ellipse. Find the maximum value of △△AOB area. 3. The distance from the endpoint to the left and right focus is equal (the definition of ellipse), so we can know that a=√3, c/a=√6/3, substitute c=√2, B = √ (A 2-C 2) = 1, and the equation is X 2/3+Y 2/65438. The solution of y=x+ 1 gives x 1 = 0, y 1 = 1, x2 =- 1.5, y2 =-0.5. The formula of chord length is √ (1+k 2) [Assuming that we have found the parallel line with the longest distance from p to chord, we can find that the parallel line is the tangent of ellipse and the tangent chord is parallel, then the slope of the slope chord is =, let y=x+m, and use the discriminant to equal 0 to get m=2, -2. Combined with the graph, we can get m =-2. x = 1.5，y =-0。 The trilinear equation x-y+ 1=0 is obtained by the distance formula from point to line, and the area is 1/2 * √2/2 * 3 √ 2/2 = 3/4.

hyperbola

Definition: We call the locus in which the absolute value F 1 and F2 of the distance difference between two fixed points in a plane is equal to a constant a hyperbola. Definition 1:

In a plane, the locus of a point whose absolute value is constant (less than the distance between two fixed points [1]) is called hyperbola. Definition 2: On the plane, the locus of a point whose distance ratio from a given point to a straight line is a constant greater than 1 is called hyperbola. Definition 3: Plane cutting conical surface. When the section is not parallel to the generatrix of the conical surface and intersects with the two conical surfaces of the conical surface, the intersection line is called hyperbola. Definition 4: In the plane rectangular coordinate system, when the binary quadratic equation f (x, y) = ax 2+bxy+cy 2+dx+ey+f = 0 meets the following conditions, it is hyperbolic. 1.a, b and c are not all zero. 2.B 2-4ac > 0。 In analytic geometry in senior high school, we learned that the center of a hyperbola is at the origin, like it is symmetrical about X and Y. At this point, the hyperbolic equation degenerates into: x 2/a 2-y 2/b 2 = 1. The above four definitions are equivalent.

Simple geometric properties of hyperbola

1, the value range of a point on the trajectory: │x│≥a (focusing on the X axis) or │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 the real axis of hyperbola and │AA'│=2a. B(0，-b)，B'(0，B)。 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 that distance from the focus to the directrix and θ is the angle between the chord and the X axis. Let 1-ecosθ=0 to find θ, where θ 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 the hyperbola center) x = [(EP/1-e)+(-EP/1+e)]/2 (please simplify) The straight line ρ cos θ = [(EP/1-e)+(-EP/). Rotate the line clockwise by π/2-arccos( 1/e) to get the asymptote equation. Let the rotated angle be θ′, then θ′ = θ-[pi/2-arccos (1/e)], then θ = θ′+[pi/2-arccos (1/e)] is substituted into the above formula: ρ cos {θ′+[pi/e]. 2 namely: ρ sin [arccos (1/e)-θ′] = [(EP/1-e)+(-EP/1+e)]/2 Now you can use θ instead of θ′ in the formula to get the equation: ρ sin [arccos (+e)] It is proved that the point on the hyperbola x 2/a 2-y/b 2 =1is the point on the first quadrant of the asymptote, so y = (b/a) √ (x 2-a 2) (x > a) is because x 2-a 2.

Edit this paragraph, hyperbola focus triangle area formula

If ∠F 1PF2=θ, then s △ f 1pf2 = b 2 * cot (θ/2) or s △ f 1pf2 = b 2 */tan (θ/2). For example: the known f65438+. Solution: S △ F 1pF2 = b 2 * COT (θ/2) = √ 3 is obtained from the hyperbolic focal triangle area formula. Let the distance between P and X axis be h, then s △ f1pf 2 =1/2 * h * 2 √ 2; h=√6/2

Edit this paragraph, hyperbolic parametric equation

Parametric equation of hyperbola: x=a*sec θ (secθ) y=b*tan θ (a is the real semi-axis length, b is the imaginary semi-axis length, and θ is the parameter. )