It is known that the center of the ellipse C is at the origin, the focus is on the X axis, the eccentricity is 1 / 2, and one of its vertices is just the focus of the parabola x ^ 2 = - 12Y

It is known that the center of the ellipse C is at the origin, the focus is on the X axis, the eccentricity is 1 / 2, and one of its vertices is just the focus of the parabola x ^ 2 = - 12Y

x²=-12y
2p=12
p/2=3
So the focus is (0, - 3)
That is, B = 3
e²=c²/a²=1/4
c²=a²/4
Then B & # 178; = 9 = A & # 178; - A & # 178; / 4
a²=12
x²/12+y²/9=1
It is known that the center of the ellipse C is the origin of the rectangular coordinate system xoy, the focus is on the x-axis, and the distances from one vertex to two focuses are 7 and 1 respectively
It is known that the center of the ellipse C is the origin of the rectangular coordinate system xoy, the focus is on the X axis, and the distances from one vertex to two focuses are 7 and 1, respectively
To solve the elliptic equation: (1);
(2) If P is a moving point on the ellipse C, M is a point on a straight line passing through P and perpendicular to the x-axis, iopi = λ iopi, find the trajectory equation of point m, and explain what kind of curve the trajectory is
Do you have a question in your second question? Iopi = λ iopi? Is it not true? It should be iopi = λ iomi, and λ is the eccentricity of ellipse? If it is, this question should be the one in 2009 Ningxia Hainan paper. (I) let the length of the long half axis of ellipse and be a, C respectively. From the known a-c = 1, a + C = 7, the solution is a = 4, C = 3
1) If the focus is on the x-axis, and the distance between a vertex and two focuses is 7 and 1 respectively, then the vertex should be on the x-axis, and the focal length = 7-1 = 6. Suppose the focal coordinates F1 (- C, 0), F2 (C, 0), C = 6 / 2 = 3, the long half axis a = C + 1 = 4, the short half axis B = √ (a ^ 2-C ^ 2) = √ 7, and the elliptic equation is: x ^ 2 / 16 + y ^ 2 / 7 = 1
(2) | op | / | om | = λ, let m (x, y), P (x, K), P point be equal to abscissa of M, K is ordinate, | op | = √ (x ^ 2 + K ^ 2), | om | = √... Expand
1) If the focus is on the x-axis, and the distance between a vertex and two focuses is 7 and 1 respectively, then the vertex should be on the x-axis, and the focal length = 7-1 = 6. Suppose the focal coordinates F1 (- C, 0), F2 (C, 0), C = 6 / 2 = 3, the long half axis a = C + 1 = 4, the short half axis B = √ (a ^ 2-C ^ 2) = √ 7, and the elliptic equation is: x ^ 2 / 16 + y ^ 2 / 7 = 1
(2) | op | / | om | = λ, let m (x, y), P (x, K), P point equal to abscissa of M, K is ordinate, | op | = √ (x ^ 2 + K ^ 2), | om | = √ (x ^ 2 + y ^ 2), P on ellipse, x ^ 2 / 16 + K ^ 2 / 7 = 1,
K = √ 112-7x ^ 2) / 4, x ^ 2 + (112-7x ^ 2) / 16 = λ ^ 2 (x ^ 2 + y ^ 2)
x^2/(7(16λ^2-9)/16λ^2+y^2/7=1
When λ / 3 is ellipse, λ > 4
It is known that the center of the ellipse C is the origin of the rectangular coordinate system xoy, the focus is on the x-axis, and the distances from one vertex to two focuses are 7 and 1 respectively. (1) find the equation of the ellipse C; (2) if P is a moving point on the ellipse C, M is a point on a line passing through P and perpendicular to the x-axis, | op ||| om | = λ, find the trajectory equation of the point m, and explain what curve the trajectory is
(1) Let a − C = 1A + C = 7, a = 4, C = 3, so the equation of ellipse C is x216 + Y27 = 1. (2) let m (x, y), where x ∈ [- 4, 4]. From the known | op | 2 | om | 2 = λ 2 and point P on ellipse C, we can get 9x2 + 11216 (x2 + Y2) = λ 2, and sort out (16 λ 2-9) x2 + 16 λ 2Y2 = 112, where x ∈ [- 4, 4]. ① when λ = 34, we can get 9 Therefore, the trajectory equation of point m is y = ± 473 (- 4 ≤ x ≤ 4), and the trajectory is two line segments parallel to the X axis. ② when λ≠ 34, the deformation of the equation is x211216 λ 2 − 9 + y212116 λ 2 = 1, where x ∈ [- 4, 4]; when 0 ＜λ＜ 34, the trajectory of point m is the part of hyperbola with the center at the origin and the real axis on the Y axis satisfying - 4 ≤ x ≤ 4; when 34 ＜λ＜ 1, the trajectory of point m is the part with the center at the origin and the real axis on the Y axis satisfying - 4 ≤ x ≤ 4 When λ≥ 1, the locus of point m is an ellipse with the center at the origin and the major axis on the X axis
It is known that the center of ellipse C is the origin of rectangular coordinate system, and the focus is on the X axis. The distances from one vertex to two focuses are 7 and 1 respectively. The center of ellipse C is known
It is known that the center of the ellipse C is the origin of the rectangular coordinate system xoy, the focus is on the s-axis, and the distances from one vertex to two focuses are 7 and 1, respectively
It is known that the center of the ellipse C is the origin of the rectangular coordinate system xoy, the focus is on the s-axis, and the distances from one vertex to two focuses are 7 and 1, respectively
If P is a moving point on the ellipse C, M is a point on a straight line passing through P and perpendicular to the X axis, ∣ op ∣ / ∣ om ∣ = e, find the trajectory equation of point m, and explain what kind of curve the trajectory is
ellipse.
If sin (π / 4 + x) = 1 / 3, then sin2x is equal to
sin(π/4+x)=-3/5
sinxcosπ/4+cosxsinπ/4=-3/5
√2/2(sinx+cosx)=-3/5
sinx+cosx = -3√2/5
(sinx+cosx)^2 = (-3√2/5)^2
sin^2x+cos^2x+2sinxcosx=18/25
1+sin2x=18/25
sin2x = -7/25
Positive scale function y = KX, why K is not equal to 0
Why?
If K is equal to 0, the function becomes y = 0, which is a constant function, not a positive proportion function
If k = 0, then the value of Y is always 0, which has nothing to do with the value of X. it is not a positive proportional function
It is proved that the distance from any point to the center of an equiaxed hyperbola is the proportional median of its distance from two focal points
Hyperbola assumes that x ^ 2-y ^ 2 = a ^ 2, let x = ASEC *, y = atan * (* as parameter) the square of the distance from the point to the center be a ^ 2 [(SEC *) ^ 2 + (Tan *) ^ 2] to the two focal points √ a ^ 2 [(SEC * + √ 2) ^ 2 + (Tan *) ^ 2] and √ a ^ 2 [(SEC * - √ 2) ^ 2 + (Tan *) ^ 2] respectively, as long as it is proved that the product of the latter two formulas is the first formula
Given sin (x-quarter) = 3 / 5, what is the value of sin2x
Sin (x - π / 4) = 3 / 5sinx · cos π - cosx · sin π = 3 / 5 √ 2 / 2 SiNx - √ 2 / 2 cosx = 3 / 5 from the above formula: SiNx cosx = 3 / 5 * √ 2 square both ends of the equation, we get: (cosx) ^ 2 + (SiNx) ^ 2 - 2sinx cosx = 9 / 25 * 21-sin2x = 18 / 25, so sin2x = 7 / 25
It's 7 / 25
First, expand the left formula, sinxcos π / 4-cosxsin π / 4 = 3 / 5, extract two-thirds of the root sign, and get the root sign two of SiNx cosx = 3 / 5 times, and then expand the square of both sides to get the result. The premise is that you know sin2x = 2sinxcosx
The image with positive scale function y = KX (k is not equal to 0) is a straight line passing through ()
What are the brackets for?
The image with positive scale function y = KX (k is not equal to 0) is a straight line passing through (origin)
Or the image with positive scale function y = KX (k is not equal to 0) is a straight line passing through (0,0)
Passing through the center point and origin
It's a straight line
Find the standard equation of hyperbola suitable for the following conditions: (1) the focus is on the x-axis, a = 4, B = 3
(2) the focus is on the x-axis, passing through the points (- radical 2, - radical 3), (3 / 3 radical 15, radical 2); and (3) the focus is (0, - 6), (0,6) and passing through the points (2,
Question 1: x ^ 2 / 16-y ^ 2 / 9 = 1
Question 2: x ^ 2-y ^ 2 / 3 = 1
Question 3: y ^ 2 / 20-x ^ 2 / 16 = 1
The focus is on the X axis, where x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1. The first question can be replaced by AB, and the second question can be replaced by (- radical 2, - radical 3), (root 15 of 3, radical 2).
(3) Let y ^ 2 / A ^ 2-x ^ 2 / b ^ 2 = 1, a ^ 2-B ^ 2 = 36 and then bring in the point (2, - 5) to get the solution of 25 / A ^ 2-4 / b ^ 2 = 1··········