If the ellipse C1: + = 1 (0

If the ellipse C1: + = 1 (0

This is the derivation of parabolic equation, the purpose is to find the slope of tangent
The equation is y = 1 / 4x ^ 2, and the derivative is y '= 1 / 4 * 2x = x / 2
So the slope of tangent is X1 / 2 and X2 / 2 respectively
It is known that the focus of ellipse C1 and hyperbola C2 is F1 (- radical 2,0) F2 (radical 2,0), and a common point of C1 and C2 is p (radical 2.1)
(1) The equation for solving ellipse C1 and hyperbola C2
(2) The equation for finding the tangent of hyperbola C2 passing through point m (0,2)
One
Definition by ellipse: C ^ 2 = 2 = a ^ 2 - B ^ 2
[x ^ 2 / A ^ 2] + [y ^ 2 / (a ^ 2 - 2)] = 1 (√ 2, 1)
The solution is: A ^ 2 = 1 or 4, ∵ a ^ 2 > C ^ 2 = 2, ∵ a ^ 2 = 4, B ^ 2 = 2
The elliptic equation: [x ^ 2 / 4] + [y ^ 2 / 2] = 1
According to hyperbola definition: C ^ 2 = a ^ 2 + B ^ 2 = 2
[x ^ 2 / A ^ 2] - [y ^ 2 / (2 - A ^ 2)] = 1 (√ 2, 1)
The solution is: A ^ 2 = 1 or 4, ∵ a ^ 2 < C ^ 2 = 2, ∵ a ^ 2 = 1, B ^ 2 = 1
The hyperbolic equation: x ^ 2 - y ^ 2 = 1
Two
Let the straight line passing through M (0,2) be y = KX + 2, and the simultaneous hyperbolic equation be y = KX + 2
(kx + 2)^2 = x^2 - 1 ,∴(k^2 - 1)x^2 + 4kx + 5 = 0 ,
∵ the asymptote of hyperbola is y = x and y = - X. therefore, if K ^ 2 - 1 = 0 (i.e. k = 1 or - 1), the straight line is parallel to the asymptote, not tangent to the hyperbola and has an intersection, ∵ K ^ 2 ≠ 1,
Therefore, taking △ as 0, we can get: 16K ^ 2 = 20 (k ^ 2 - 1), K ^ 2 = 5,
Ψ k = √ 5 or - √ 5
There are two tangent equations satisfying the conditions
L1 : y = √5x + 2
L2 : y = -√5x + 2
The shortest distance from the point on ellipse C1: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 to the directrix of parabola C2: x ^ 2 = 6By is 1 / 2, and the eccentricity of ellipse C1 is root sign 3 / 2
Let the right focus of C1 be e, the focus of C2 be f, and point p be the moving point on C2. If the area of triangle EFP is m, how many points P are there
As far as I know, we need to discuss it by classification. Let's discuss it according to the size of M. first solve it, C1; X ^ 2 / 4 + y ^ 2 / 1 = 1, C2: x ^ 2 = 6y, so: F (0,1.5) e (radical 3,0) let P (x, x ^ 2 / 6) bring into EF straight line: X / √ 3 + 2Y / 3 = 1. Using the distance formula, we can get: D = ∣ X / √ 3 + x ^ 2 / 6 × 2 / 3-1 ∣ / √ (1 / 3 + 4 / 9)
If the parabola y = - X & # 178; - 2x + C passes through the origin and the vertex is a, then the analytic expression of the positive proportion function passing through a is
y=-x^2-2x-1+(1+c)=-(x-1)^2+(1+c),
So, x = 1, the highest point is (1 + C)
Positive scale function: y = (1 + C) x
Given that the equation X22 + m − y2m + 1 = 1 represents hyperbola, then the value range of M is ()
A. M ＜ 2B. 1 ＜ m ＜ 2C. M ＜ - 2 or m ＞ - 1D. M ＜ - 1 or 1 ＜ m ＜ 2
∵ X22 + m-y2m + 1 = 1 means hyperbola, ∵ (2 + m) (M + 1) > 0. The solution is m ＜ - 2 or m ＞ - 1. The value range of ∵ m is m ＜ - 2 or m ＞ - 1
It is known that log2sin (3-A) = - 2, and Tana
Log2sin (3 π - a) = - 2, that is sin (3 π - a) = 1 / 4 = sin (5 π - a) = sin (- 5 π - a) = - sin (5 π + a) = 1 / 4
That is sin (5 π + a) = - 1 / 4
Because of Tana
Among the following functions, a, y = x / 2 B, y = - (2 / x) C, y = - [(x-1) / 2] d, y = (X & # 178; - 1) / X are linear functions but not positive proportional functions
Tell me why I chose this
C
Given that the equation X22 + m − y2m + 1 = 1 represents hyperbola, then the value range of M is ()
A. M ＜ 2B. 1 ＜ m ＜ 2C. M ＜ - 2 or m ＞ - 1D. M ＜ - 1 or 1 ＜ m ＜ 2
∵ X22 + m-y2m + 1 = 1 means hyperbola, ∵ (2 + m) (M + 1) > 0. The solution is m ＜ - 2 or m ＞ - 1. The value range of ∵ m is m ＜ - 2 or m ＞ - 1
Sina and cosa are the third quadrant. Which quadrant is Tana / 2 in (alpha)
The second or fourth quadrant
In function 1. Y = - 5x + 12. Y = - 4 / X3. Y = 3x & # 178; 4. Y = 1 / 2x5. Y = 5 + 2x______ Is a positive proportional function______ .
In function
y=-5x+1
y=-4/x
y=3x²
Y = 1 / 2x & nbsp; (here is equivalent to y = 0.5x? Not y = 1 / (2x))
In y = 5 + 2x,
Is a function of degree_ 1,4,5_____ Is a positive proportional function__ 4____ .