![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
How to determine the axis of the focus of the ellipse How to determine the axis of focus of ellipse, hyperbola and parabola?
Ellipse: X & sup2 / / A & sup2; + Y & sup2 / / B & sup2; = 1 (a, b > 0) compare a and B, which is bigger on which axis;
Example: X & sup2 / 3 + Y & sup2 / 2 = 1 on the x-axis, X & sup2 / 3 + Y & sup2 / 12 = 1 on the y-axis
Hyperbola: X & sup2; / A & sup2; - Y & sup2; / B & sup2; = 1 or Y & sup2; / B & sup2; - X & sup2; / A & sup2; = 1, whose front coefficient is negative, on whose axis
For example: X & sup2 / 5-y & sup2 / 4 = - 1, which becomes the standard formula, - X & sup2 / 5 + Y & sup2 / 4 = 1, on the x-axis
Parabola: Y & sup2; = 2px, or X & sup2; = 2PY, who is once, whose axis is the focus,
For example: Y & sup2; = 2px on the x-axis,
The earth's orbit is an ellipse. One focus is the sun. What's the other focus?
Is the earth's orbit influenced by other planets to form an ellipse? Is the orbit of the moon, the only natural satellite of the earth, also an ellipse? Why?
Newton thought it was caused by gravitation. Einstein thought it was an object with enough mass to distort its surrounding space, but to move around its star. This is a superposition effect, just like the orbits of many places with gravitational balance between planets are interlinked
There is an elliptical comet orbit map, which is 4 in length, 2 in height and 3 in root sign. It is known that point O is the center of the ellipse, A1A2 is the two ends of the long axis, and the sun is at the left focus F of the ellipse. (1) establish an appropriate coordinate system, write out the elliptic equation, and find the distance between the two when the comet moves directly above the sun. (2) the line L is perpendicular to the extension line of A1A2 and at point D, | OD | = 4, A2p intersects the ellipse at two points m and n (different from A1 and A2). Can point A2 be on a circle with diameter Mn? Explain the reason
(1) Establish the coordinate system as shown in the figure, let the elliptic equation be + = 1 (a > b > 0), according to the meaning, 2A = 4, 2b = 2, ∧ a = 2, B =. ∧ C = 1. The elliptic equation is = 1, f (- 1,0), substituting x = - 1 into the elliptic equation to get y = ±, ∧ when the comet is directly above the sun, the distance between them in the figure is 1.5 cm. (2) from (1), a 1 (-)
The two focus coordinates of the ellipse are F1 (- radical 3,0) (radical 3,0), and the ellipse passes (1, - radical 3 / 2)
The two focus coordinates of the ellipse are F1 (- radical 3,0) (radical 3,0), and the ellipse passes (1, - radical 3 / 2)
(1) Solving elliptic equation
(2) Cross the point (- 6 / 5,0) and make a straight line L not perpendicular to the y-axis. The ellipse is at two points m and n. A is the left vertex of the ellipse. Try to judge whether the size of ∠ man is a fixed value and explain the reason
In fact, I just need to help me to answer the second question. I see the answer: let x = KY-6 / 5 be the equation of simultaneous ellipse and straight line. Weida's theorem is y1y2, Y1 + Y2, and the final point multiplies the vector am an = (x1 + 2, Y1) (X2 + 2, Y2) = (k ^ 2 + 1) y1y2 + 4 / 5 (Y1 + Y2) + 16 / 25 = 0
AM AN =(x1+2,y1)(x2+2,y2)
=x1x2+2(x1+x2)+y1y2+4
M. If n is on the straight line x = KY-6 / 5, then:
x1=ky1-6/5
x2=ky2-6/5
The results are: x1x2 = K & # 178; y1y2 - (6 / 5) K (Y1 + Y2) + 36 / 25
x1+x2=k(y1+y2)-12/5
So, am an = x1x2 + 2 (x1 + x2) + y1y2 + 4
=(k²+1)y1y2+(4/5)k(y1+y2)+16/25
PS: you missed a k after 4 / 5
It is known that the two focal points of the ellipse are F1 (- √ 3,0), F2 (√ 3,0), and the eccentricity e = √ 3 / 2
If l and the ellipse intersect at P and Q, and | PQ | is equal to the length of the minor axis of the ellipse, find the value of M
C = √ 3, a = 2, B = 1, elliptic equation: X & # 178; + 4 + Y & # 178; = 1, combined with linear equation: y = x + m, eliminate y to get: 5x & # 178; + 8mx + 4m & # 178; - 4 = 0
X1 + x2 = - 8m / 5 X1 * x2 = (4m & # 178; - 4) / 5 m = + - 30 / 4 is obtained by solving the chord length formula with √ 2 * (x1-x2) = 2
If there is a point A / sinpf1f2 = C / sinpf2f1 on the ellipse, the range of eccentricity of the ellipse is?
a/sinPF1F2=c/sinPF2F1
c/a=sinPF2F1/sinPF1F2
According to the sine theorem, sinpf2f1 / sinpf1f2 = | PF2 | / | Pf1|
So, e = C / a = | PF2 | / | Pf1|
|PF1|+|PF2|=2a
So, (E + 1) | Pf1 | = 2A
|PF1|=2a/(e+1)
|PF2|=e|PF1|=2ae/(e+1)
And: | Pf1 | - | PF2 | ≤| F1F2 | = 2C
So. 2A (1-e) / (E + 1) ≤ 2C
(1-e)/(1+e)≤e
e^2+2e-1≥0,e>0
Therefore, e ≥ √ 2-1
The range of ellipse eccentricity is: [√ 2-1,1)
The two focuses of ellipse y ^ 2 / A ^ 2 + x ^ 2 / b ^ 2 = 1 (a > b > 0) are F1 (0, - C)), F2 (0, c) (c > 0), and the eccentricity e = √ 3 / 2,
The elliptic equation is: Y & # 178 / 4 + X & # 178; = 1
Let PQ be the intersection of the ellipse and the straight line y = x + 1, and find the value of Tan ∠ poq
Substituting x = Y-1 into the ellipse, we get: Y & # 178; / 4 + (Y-1) &# 178; = 1
5y²-8y=0
y1=0,y2=8/5
Then: X1 = - 1, X2 = 3 / 5
So, P (- 1,0), q (3 / 5,8 / 5)
Drawing: Tan ∠ poq = - Tan ∠ qox = - K (OQ) = - 8 / 3
If the focal length, minor axis length and major axis length of an ellipse form an arithmetic sequence, then the eccentricity is______ ．
According to the meaning of the question, the focal length, the length of the minor axis and the length of the major axis of the ellipse form a sequence of arithmetic numbers, and the answer is: 35
If the length of the major axis, the length of the minor axis and the focal distance of an ellipse form an arithmetic sequence, then the eccentricity of the ellipse is zero______ ．
According to the meaning of the question, 2a, 2b and 2c are equal difference sequence 2B = a + C 4B2 = A2 + 2Ac + C2 B2 = a2-c2 ②, 5c2 + 2ac-3a2 = 0 e = Ca 5e2 + 2e-3 = 0 0 e 1 e = 35, so the answer is: 35
The eccentricity of an ellipse is ()
A. 45B. 35C. 25D. 15
Let the major axis be 2a, the minor axis be 2B, and the focal length be 2c, then 2A + 2C = 2 × 2B, that is, a + C = 2B {(a + C) 2 = 4B2 = 4 (a2-c2), so 3a2-5c2 = 2Ac, divide A2, and get 5e2 + 2e-3 = 0, ∧ e = 35 or E = - 1 (rounding off), so select B
RELATED INFORMATIONS
- 1. It is known that the equation of ellipse C is 3xx + 4yy = 12, and the value range of M is obtained, so that two different points on the ellipse are symmetrical about the line This line: y = 4x + M I'm sorry I knocked it out
- 2. The chord length of the line y = X-1 / 2 cut by the ellipse x ^ 2 + 4Y ^ 2 = 4 is
- 3. It is known that the ellipse X & # 178; / A & # 178; + Y & # 178; / B & # 178; = 1 (a ＞ 0b ＞ 0) and the hyperbola X & # 178; - Y & # 178; / 2 = 1 have the same focus. F1. F2. P is a common intersection of the hyperbola and the ellipse, and | Pf1 |. | PF2 | = 3 - the equation for finding the ellipse, if the straight line y = x + m intersects the ellipse and a, B, two points, O are the coordinate origin, try to find the maximum area of the triangle AOB
- 4. It is known that tan2x = - 2 √ 2, π - 2
- 5. If 45 ＜ x ＜ 90, then the maximum value of function y = tan2x (TaNx) to the third power is?
- 6. TaNx = 1,3sinb = sin (2x + b), find Tan (x + b)
- 7. Given sin (π + x) = 4 / 5, X ∈ (π / 2,3 π / 2), then TaNx=
- 8. TaNx = 5 / 12, tany = 4 / 3. X belongs to 0,90, y belongs to 180270, find sin (x + y)
- 9. If x belongs to (0, π / 2), find the minimum value of 2tanx + 1 / TaNx
- 10. F (x) = 1 + TaNx / 1 + (TaNx) ^ 2, X belongs to [Pai / 12, Pai / 2], find the value range of F (x)
- 11. Taking the moving points P, F1 and F2 on the ellipse x ^ 2 / 25 + y ^ 2 / 16 = 1 as the focus, the trajectory equation of the center of gravity g of △ pf1f is obtained
- 12. If a straight line passing through the left focus of the ellipse x ^ 2 + 2Y ^ 2 = 4 with a left inclination angle of 30 degrees intersects the ellipse at two points a and B, then the chord length AB = 16 / 5
- 13. The left and right focus of hyperbola F1F2, x ^ 2-y ^ 2 / 9 = 1, point P on hyperbola, vector Pf1 * PF2 = 0, find the absolute value of vector Pf1 + PF2
- 14. There is no intersection between the circle whose diameter is the chord of the conic curve passing through the focus and the corresponding guide line. What is the conic curve 1. Ellipse 2. Hyperbola 3. Parabola 4. Not sure
- 15. The elliptic equation with focus on x-axis, focal length of 4 and eccentricity of 1 / 2 is
- 16. It is known that the focus of the ellipse C is on the x-axis, the eccentricity e = 3 / * 3, and the area of the quadrilateral with four vertex structure s = 2 * 6 It is known that the focus of the ellipse C is on the x-axis, the eccentricity e = 3 / * 3, and the area s of the quadrilateral with four vertex structure is 2 * 6. The standard equation for finding the ellipse C is 〈 * means root 〉
- 17. Given that the eccentricity of ellipse y ^ 2 / A ^ 2 + x ^ 2 / b ^ 2 = 1 (a > b > 0) is 3 / 5, and the distance between two focal points is 3, the value of a + B is obtained
- 18. If the focus coordinates of the ellipse are (- 5,0) and (5,0), and the sum of the distances between a point on the ellipse and two focuses is 26, then the equation of the ellipse is () A. x2169+y2144=1B. x2144+y2169=1C. x2169+y225=1D. x2144+y225=1
- 19. The ellipse equation is x 2A2 + y 2B2 = 1 (a > b > 0), a vertex is a (0,2), eccentricity e = 63. (1) the equation for solving ellipse; (2) the line L: y = kx-2 The ellipse equation is x2 / A2 + Y2 / B2 = 1 (a > b > 0), a vertex is a (0,2), and the eccentricity e = root 6 / 3 (1) Find the equation of ellipse; (2) The straight line L: y = kx-2 (K ≠ 0) intersects the ellipse at two different points m, n satisfies MP → = PN →, AP → &; Mn → = 0, find K
- 20. If f (x) = sin (2x-a) + 1 satisfies f ((π / 3) + x) = f ((π / 3) - x, what is a equal to?