Given that K is a real number, if the focal length of hyperbola X2K − 5 + Y22 − | K | = 1 has nothing to do with the value of K, then the value range of K is () A. （-2，0]B. （-2，0）∪（0，2）C. [0，2）D. [-1，0）∪（0，2]

Given that K is a real number, if the focal length of hyperbola X2K − 5 + Y22 − | K | = 1 has nothing to do with the value of K, then the value range of K is () A. （-2，0]B. （-2，0）∪（0，2）C. [0，2）D. [-1，0）∪（0，2]

The equation is hyperbola ⇔ (K-5) (2 - | K |) < 0 ⇔ - 2 ＜ K ≤ 0 or 0 ＜ K ＜ 2 or K ＞ 5; when - 2 ＜ K ≤ 0, the equation is Y22 + K − X25 − k = 1, A2 = 2 + K, B2 = 5-K, then C2 = 7 is independent of K; when 0 ＜ K ＜ 2, the equation is Y22 + K − X25 − k = 1, A2 = 2-k, B2 = 5-K, then C2 = 7-2k and K

Given that a focal point of the ellipse is f (1,0), the corresponding collimator is x = 2, and the eccentricity is √ 2 / 2, the equation of the ellipse is obtained

It is known that a focal point of an ellipse is f (1,0), and the corresponding directrix is x = 2
a²/c=2
The eccentricity of ellipse is √ 2 / 2
∴e=c/a=√2/2
∴a=√2,c=1,a²=2
∴b²=a²-c²=1
The equation of ellipse is X & # 178 / 2 + Y & # 178; = 1

The equation of the circle with the left focus of the ellipse as the center and the length of the long half axis as the radius?

If the focus of the ellipse is on the x-axis, then the left focus is (- C, 0), and the length of the major half axis is a, then the equation of the circle is (x + C) ^ 2 + y ^ 2 = a ^ 2

If the focus of the ellipse is on the y-axis, the eccentricity is 1 / 3, and the distance between the two focuses is 6, then the equation of the ellipse is

E = C / a = 1 / 3, then C ^ 2 / A ^ 2 = (a ^ 2-B ^ 2) / A ^ 2 = 1 / 9, -- (1)
And 2C = 6, so C = 3, -- (2)
The above results show that a ^ 2 = 81, B ^ 2 = 72,
So the elliptic equation is y ^ 2 / 81 + x ^ 2 / 72 = 1

It is proved that if f (x) is a periodic function with L as period, then f (AX + b) (a, B are constants, and a > 0) is a periodic function with L / A as period

If the period of F (x) is I, then according to the definition, f (x + Ki) = f (x)
If y = x + Ki, then f (y) = f (x);
For the function: G (x) = f (AX + b), the,
When y = x + K (I / a), ay + B = a [x + K (I / a)] + B = ax + B + ki
g(y)=f(ay+b)=f(ax+b+kI)
According to the periodic property of F (x), f (AX + B + Ki) = f (AX + b) = g (x)
So there are:
When y = x + K (I / a), G (y) = g (x)
That is g (x + K (I / a)) = g (x)
So g (x) = f (AX + b) is a periodic function with I / a period

Given the standard equation of hyperbola, how to find its asymptote equation. Give an example

The best way to remember is: change the "1" on the right side of the hyperbolic standard curve equation: X & sup2; / A & sup2; - Y & sup2; / B & sup2; = 1 to "0", that is: X & sup2; / A & sup2; - Y & sup2; / B & sup2; = 0, so, Y & sup2; = B & sup2; X & sup2; / A & sup2; so, its asymptote is: y = BX / a