![](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>)
High school mathematics! Does hyperbola have a straight line?
There are two cases,
If the hyperbolic equation is x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1, then its quasilinear is x = ± a ^ 2 / C
If the hyperbolic equation is y ^ 2 / A ^ 2-x ^ 2 / b ^ 2 = 1, then its quasilinear is y = ± a ^ 2 / C
Yes...
If you have a hyperbola, you don't have a quasicollis,
The criterion line is x = ± A & sup2. / C, where the denominator C cannot be equal to 0, otherwise it is meaningless,
Because of this, the circle has no guide line
Yes, it is the same as ellipse in form.
If the eccentricity of hyperbola x2a2 − y2b2 = 1 and ellipse x2m2 + y2b2 = 1 (a > 0, M > b > 0) are reciprocal, then the triangle with a, B, m as side length is ()
A. Acute triangle B. obtuse triangle C. right triangle D. isosceles triangle
The eccentricity of hyperbola x2a2 − y2b2 = 1 and ellipse x2m2 + y2b2 = 1 (a > 0, M > b > 0) are reciprocal, so A2 + b2a2 · M2 − b2m2 = 1, so b2m2-a2b2-b4 = 0, that is M2 = A2 + B2, so the triangle with a, B, m as the side length is right triangle
Ellipse and hyperbola image problem
What are a, B, C, F1 and F2 in the elliptic and hyperbolic equations? What is their relationship? It's better to come to the image. I want to know which segment a is and which segment B is. It's more intuitive to find the image,
Ellipse A is the major axis, B is the minor axis, C is the focal length. A ^ 2 = B ^ 2 + C ^ 2
F1, F2 are the focus
Hyperbola a a is real axis, B is imaginary axis, C is focal length. C ^ 2 = a ^ 2 + B ^ 2
F1, F2 are the focus
Ask the great God to help me solve the elliptic and hyperbolic equations in detail a, B, C, F1, F2 are what? What's the relationship between them? I want to know which segment a is and which segment B is? For images, more intuitive, satisfactory bonus.
It's all parameters
Which line is the Quasilinear of ellipse and hyperbola?
I still can't understand it. I don't want to know its expression. What I want to know is how to define the Quasilinear of ellipse and hyperbola
2. Unified definition of this curve
The locus of a point in the plane whose ratio of distance to fixed point F to fixed line L is constant E
Here f is the focal point, l is the directrix, and E is the eccentricity
Both ellipse and hyperbola have two quasars corresponding to their respective focal points
If the image of the first-order function y = (M-3) x + m + 1 passes through the first, second and fourth quadrants, then the value range of M is______ .
∵ the image of the first-order function y = (M-3) x + m + 1 passes through the first, second and fourth quadrants, ∵ m − 3 ＜ 0m + 1 ＞ 0, and the solution is - 1 ＜ m ＜ 3
Given the equation of circle x2 + Y2 + KX + 2Y + K2 = 0 (k is a real number), if the fixed point a (1,2) is outside the circle, find the value range of K
If (1,2) is brought in so that the result is greater than 0, then the range of K is any real number
Then the equation is transformed into the standard form (x + K / 2) ^ 2 + (y + 1) ^ 2 = 1-3 / 4K ^ 2, and the right end of the equation should be greater than 0
Get the results
Judge (x ^ 2 + 3sinx sin / 4) '= 2x + 3cosx CoS / 4
Wrong
Sin π / 4 is a constant and the derivative is 0
So it should be 2x + 3cosx
No, = 2x + 3cosx
Sin / 4 is a constant, 0
No, it should be 2x + 3cosx, sin / 4 is a constant, and the derivative of the constant is 0
(x ^ 2 + 3sinx sin / 4) '= 2x + 3cosx
Sin / 4 is a constant and the derivative of the constant is 0
If the image of linear function y = MX - (4m-4) passes through the first, second and third quadrants, try to find the value range of M
Because the image of the linear function y = MX - (4m-4) passes through the first and third quadrants,
So m > 0,
And through the second quadrant,
So - (4m-4) > 0,
So m
The slope of the solution is greater than 0-4m and the intercept is greater than 0-4m
Firstly, the image of the first-order function y = MX - (4m-4) passes through the first and third quadrants, so m > 0;
Secondly, the image passes through the second quadrant, so - (4m-4) > 0, M is obtained
The line y = KX + 1 and the hyperbola 3x (square) - Y (square) = 1 intersect at two points a and B. what is the value of K when two points a and B are on the same branch of the hyperbola（
Suppose a (x1, Y1) B (X2, Y2)
A. B two points are on the same branch,
Then x1x2 > 0, y1y2 > 0
K = 0, does not meet the requirements
k≠0
Y = KX + 1 into hyperbola 3x (square) - Y (square) = 1
(3-K^2)X^2-2KX-2=0
Discrimination > 0
4k^2+8(3-k^2)>0,-√6
The image of the function y = cos (x - π / 2) + sin (π / 3-x) is symmetric
Y = cos (x - π / 2) + sin (π / 3-x) = 3x / 2-sinx / 2 = sin (x + π / 3) SiNx is symmetric about x = k π + π / 2, so sin (x + π / 3) is symmetric about x + π / 3 = k π + π / 2, x = k π + π / 6) so y = cos (x - π / 2) + sin (π / 3-x) is symmetric about x = k π + π / 6
RELATED INFORMATIONS
- 1. Know that elliptic equation, hyperbola and it have common focus, eccentricity reciprocal, find hyperbola formula? X ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1. If you use a / B to express the equation of open hyperbola, you don't need to calculate it later, and you don't want to deduce it yourself?
- 2. It is known that the equation of ellipse C1 is x24 + y2 = 1, the left and right focus of hyperbola C2 are the left and right vertex of C1 respectively, and the left and right vertex of C2 are the left and right focus of C1 respectively
- 3. AB is the chord passing through a focus of the ellipse x ^ 2 / 5 + y ^ 2 / 4 = 1. If the inclination angle of AB is π / 3, find the chord length of ab
- 4. F1 and F2 are the two focal points of ellipse X & # 178 / A & # 178; + Y & # 178 / B & # 178; = 1 (a > b > 0). The chord AB passing through F1 and F2 form an isosceles right angle three What is the eccentricity of the ellipse when the angle BaF2 is 90
- 5. Through the focus of the ellipse X225 + Y29 = 1, the length of the chord AB with an inclination angle of 45 ° is______ .
- 6. F1 and F2 are the two focal points of the ellipse x ^ 2 / 2 + y ^ 2 = 1. If the chord AB with an inclination angle of 45 degrees is made through F2, what is the area of the triangle f1ab?
- 7. (1) Let AB be the chord passing through the center of the ellipse x * 2 / A * 2 + y * 2 / b * 2 = 1 (a > b > 0), and the left focus of the ellipse is F1 (- C, 0), then Δ F (1) Let AB be the chord passing through the center of the ellipse x * 2 / A * 2 + y * 2 / b * 2 = 1 (a > b > 0), and the left focus of the ellipse is F1 (- C, 0), then the area of Δ f1ab is the largest——————
- 8. 4X ^ 2 + 5Y ^ 2 = 1 elliptic problem The problem is 4x ^ 2 + 5Y ^ 2 = 1 to find the length of major axis and minor axis of ellipse focus
- 9. It is known that vertex BC of triangle ABC is on ellipse x ^ 2 / 3 + y ^ 2 = 1, vertex A is one focus of ellipse, and the other focus of ellipse is on edge BC to find triangle ab Girth of
- 10. Given that the vertices B and C of △ ABC are on the ellipse x23 + y2 = 1, vertex A is a focus of the ellipse, and the other focus of the ellipse is on the edge of BC, then the perimeter of △ ABC is () A. 23B. 6C. 43D. 12
- 11. If we know the intersection of the square of ellipse X / 4 + y and the square of hyperbola X-Y / 2 = 1, and F1F2 is the left and right focus of ellipse, we can find the cos angle FPF It's the focus of ellipses and hyperbolas
- 12. Eccentricity of elliptic hyperbola Literal expression of hyperbolic eccentricity
- 13. Hyperbola focuses on the focus of ellipse X / 9 + Y / 25 = 1, and its eccentricity is twice that of ellipse
- 14. Given that the eccentricity of hyperbola is 2 and has the same focus as the ellipse x ^ 2 / 25 + y ^ 2 / 9 = 1, the hypohyperbolic equation is solved
- 15. According to the conditions, the standard equation can be solved. Given that one focus of the ellipse is (0,2) and the eccentricity is 1 / 2, the standard equation can be solved
- 16. Given the coordinates of the focus of the ellipse and a point P that the curve passes through, how can we find the standard equation of the ellipse? It is known that the two focal coordinates of an ellipse are F1 (- √ 3,0) F2 (√ 3,0) and pass through the ellipse standard equation of point P (√ 5, √ 6)
- 17. It is known that the center of the ellipse e is at the origin, the focus is on the x-axis, the minimum distance between the point on the ellipse and the focus is 1, and the eccentricity e = 1 / 2 The line L: y = K (x-1) (k is not 0) intersects the ellipse e at different points P and Q (1) Solving elliptic e-equation (2) Find the range of intercept of vertical bisector of line PQ on y-axis (3) Question: is there a fixed point m on the x-axis so that the vector MP * MQ is a fixed value? If so, find out the coordinates of the fixed point m; if not, explain the reason (1) x ^ 2 / 4 + y ^ 3 = 1 has been obtained Need (2); (3) detailed explanation
- 18. It is known that the eccentricity of ellipse C is √ 6 / 3, and the distance from one end point of the minor axis to the right focus is √ 3
- 19. The eccentricity of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is √ 6 / 3, and the distance from one end point of minor axis to the right focus is √ 3 Let the intersection of line L and ellipse C and two points a and B, and the distance between coordinate origin O and line l be √ 3 / 2, then the maximum area of △ AOB can be obtained
- 20. Given that the distance from a focal point of an ellipse to the corresponding guide line is equal to the length of the major half axis of the ellipse, then the eccentricity of the ellipse is ()