The hyperbola C: 2x ^ - y ^ = 2 is known. Find the trajectory equation of the midpoint Q of the chord AB passing through the point m (2,1)

The hyperbola C: 2x ^ - y ^ = 2 is known. Find the trajectory equation of the midpoint Q of the chord AB passing through the point m (2,1)

Let a (a, b), B (C, d) Q (E, f)
2a^-b^=2 （1）
2c^-d^=2 （2）
（1）－（2）∴2（a+c）(a-c)=(b+d)(b-d)
∴k(AB)=2(a+c)/(b+d)=2e/f
Let AB line: Y-1 = K (X-2)
Bring the coordinates of Q point and K into: F (F-1) = 2e (E-2)
That is, 2x ^ - y ^ - 4x + y = 0
I don't know if it's right

If the hyperbola x ^ 2 | 4-y ^ 2 = 1 has a moving point P, O as the origin of coordinates and m as the midpoint of line OP, then the trajectory equation of point m is

Let P (a, b)
Then a ^ 2 / 4-b ^ 2 = 1
The coordinates of the midpoint of OP are [(a + 0) / 2, (B + 0) / 2]
That is, x = A / 2, y = B / 2
a=2x,b=2y
Substituting a ^ 2 / 4-b ^ 2 = 1
4x^2/4-4y^2=1
x^2-4y^2=1

Given the hyperbola 2x ^ 2-y ^ 2 = 2, find its eccentricity and asymptote equation (2)
Given the hyperbola 2x ^ 2-y ^ 2 = 2, find its eccentricity and asymptote equation (2). If a straight line intersects the hyperbola at two points a and B, and the midpoint of line AB is (2,1), find the slope of the straight line

Given the hyperbola 2x ^ 2-y ^ 2 = 2, find its eccentricity and asymptote equation (2) if a straight line intersects the hyperbola at two points a and B, and the midpoint of line AB is
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One focus of hyperbola is F1 (2, - 12), and through two points a (- 7,0), B (7,0), the trajectory equation of another focus of hyperbola is obtained~

Let the other focus of hyperbola be F2,
|A F1|=15,| B F1 |=13,
According to the definition of hyperbola, there are: || a F1 | - | a F2 || = || B F1 | - | B F2||
Remove the absolute value: | a F1 | - | a F2 | = | B F1 | - | B F2|
Or | a F1 | - | a F2 | = - (| B F1 | - | B F2 |)
When | a F1 | - | a F2 | = | B F1 | - | B F2 |,
That is 15 - | a F2 | = 13 - | B F2|
|A F2|-| B F2 |=2.
The trajectory of F2 is the right branch of hyperbola with a (- 7,0), B (7,0) as the focus, a = 1, C = 7
The trajectory equation x ^ 2-y ^ 2 / 48 = 1 (x > 0)
When | a F1 | - | a F2 | = - (| B F1 | - | B F2 |),
That is 15 - | a F2 | = - (13 - | B F2 |),
|A F2|+| B F2 |=28.
The trajectory of F2 is an ellipse with a (- 7,0), B (7,0) as the focus, a = 14, C = 7
The trajectory equation x ^ 2 / 196 + y ^ 2 / 147 = 1

Taking the opposite sides AB and CD of parallelogram ABCD as sides, make equilateral triangle Abe and equilateral triangle CDF respectively. Prove that the quadrilateral aecf is a parallelogram

It is proved that: ∵ parallelogram ABCD ∵ AB = CD, ad = BC, ∵ ABC ∵ ADC ∵ equilateral ∵ Abe ∵ be = AB = AE, ∵ Abe = 60 ∵ equilateral ∵ CDF ∵ DF = CD = CF, ∵ CDF = 60 ∵ DF = be, AE = CF ∵ CBE ∵ ABC ∵ Abe ∵ be = AB = AE,