If the mean value + 1 of the absolute value power of K of polynomial 4xy - 1 / 5 (K-2) y is a cubic binomial, then K=

If the mean value + 1 of the absolute value power of K of polynomial 4xy - 1 / 5 (K-2) y is a cubic binomial, then K=

If the polynomial 4xy ^| K | - 1 / 5 (K-2) y ^ 2 + 1 is a cubic binomial
Then 1 + | K | = 3, and K-2 ≠ 0
So k = - 2
Under the standard equation, if the ellipse passes Q (2,1) and has a common focus with x ^ / 9 + y ^ / 4 = 1, the elliptic equation can be solved
Let x ^ 2 / (9 + k) + y ^ 2 / (4 + k) = 1, substitute x = 2, y = 1 into 4 / (9 + k) + 1 / (4 + k) = 1, divide into 4 (4 + k) + (9 + k) = (9 + k) (4 + k) and simplify to K ^ 2 + 8K + 11 = 0, so K1 = (- 8 + √ 20) / 2 or K2 = (- 8 - √ 20) / 2 (rounding off). Therefore, k = (- 8 + √ 20) / 2 = √ 5-4
x^/a+y^/(a-5)=1,
Substituting (2,1), we get a = 5-radical 5 (rounding) or a = 5 + radical 5
So the equation is x ^ / (5 + root 5) + y ^ / root 5 = 1
When the function f (x) = (M2 + 2m) × XM2 + m – 1 is known, the function is an inverse proportional function
The elliptic standard equation passing through point a (- 1, - 2) and having the same focus as the ellipse x ^ 2 / 6 + y ^ 2 / 9 = 1
From the ellipse x ^ 2 / 6 + y ^ 2 / 9 = 1, we know that a ^ 2 = 9, B ^ 2 = 6, C ^ 2 = 3
Let x ^ 2 / b ^ 2 + y ^ 2 / A ^ 2 = 1, a > b > 0
Then a ^ 2-B ^ 2 = 3
Substituting point a (- 1, - 2) into x ^ 2 / b ^ 2 + y ^ 2 / A ^ 2 = 1
So 1 / b ^ 2 + 4 / A ^ 2 = 1 ②
A ^ 2 = 6 and B ^ 2 = 3 are obtained from (1) and (2)
That is, the elliptic equation is x ^ 2 / 3 + y ^ 2 / 6 = 1
Given that the function y = 2m + 10 / x-m2 + 26 is an inverse proportional function, then the value of M is ()
The denominator is the minus of X, the square of m plus 26
The expression of inverse scale function is y = K / X (k is not equal to 0)
So - m ^ 2 + 26 = 1
So m = plus or minus 5
Because 2m + 10 is not equal to 0
So m = 5
Given that the ellipse C with focus on X-axis passes through point (0,1), and the eccentricity is 2 / 3, q is the left vertex of the ellipse, the standard equation of the ellipse is solved
B = 1, C / a = 2 / 3
a=2,b=1
Elliptic standard equation x ^ 2 / 4 + y ^ 2 = 1
Inverse scale function y = (2m-1) x ^ (m ^ 2-2). When x > 0, y increases with the increase of X, and the value of M is obtained
Inverse scale function y = (2m-1) x ^ (m ^ 2-2), so there is m ^ 2-2 = - 1, that is, the solution is m = (+ / -) 1
When x > 0, y increases with the increase of X, indicating that 2m-1
A = 4. C = 3. The standard curve equation of ellipse with focus on y-axis
The straight line L with slope 1 passes through the focus of parabola y ^ 2 = 4x and intersects with parabola at two points a and B. the length | ab | of line segment AB is calculated
b^2=a^2-c^2=7
y^2/16+x^2/7=1
P=2
Focus (1,0)
Straight line y = X-1
y^2=4x=(x-1)^2
x^2-6x+1=0
|X1-x2 | = radical (△) = radical (32) = 4 radical 2
|Ab | = radical 2 | x1-x2 | = 8
If the function y = (M + 1) x ^ M2 + 2m-1 is an inverse scale function and its image is in the first and third quadrants, the value of M is obtained
If the function y = (M + 1) x ^ M2 + 2m-1 is an inverse scale function, and its image is located in the first and third quadrants, find the value of M 〈 m + 1 ＞ 0; M & # 178; + 2m-1 = - 1; (M + 1) & # 178; = 1; m + 1 = ± 1; m = 0 or M = - 2; ∵ m ＞ - 1; so m = 0; I'm glad to answer for you, skyhunter 002 will answer for you
If the two focal coordinates F1 (- 2,0) F2 (2,0) of the ellipse are known and pass P (5 / 2, - 3 / 2), then the standard equation of the ellipse is obtained
Let the standard equation of ellipse be x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1,
|F1F2|=2c=4，
So C = 2,
|PF1|+|PF2|=2a=2√10，
So a = √ 10,
b^2=a^2-c^2=6,
So the standard equation of ellipse
It is x ^ 2 / 10 + y ^ 2 / 6 = 1
How to find out Pf1 | + | PF2 | = 2A = 2 √ 10?
The two focal coordinates F1 (- 2,0) F2 (2,0) of an ellipse can be set as X & sup2; + A & sup2; + Y & sup2; = 1A & sup2; = B & sup2; + 425 / 4A & sup2; + 9 / 4B & sup2; = 1, and a & sup2; = 10, B & sup2; = 6, so the elliptic equation is X & sup2; / 10 + Y & sup2; / 6 = 1