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If the line y = KX + 1 and the ellipse X225 + y2m = 1 always have a common point, then the value range of the real number m is______ .
The straight line y = KX + 1 passes through the point (0, 1), and the straight line y = KX + 1 and the ellipse have a common point. Therefore, when (0, 1) on the ellipse or in the ellipse ﹥ 0 + 1m ≤ 1 ﹥ m ≥ 1 and M = 25, the curve is a circle or not an ellipse, so m ≠ 25. The value range of the real number m is: m ≥ 1 and m ≠ 25 & nbsp; so the answer is m ≥ 1 and m ≠ 25
If loga (2 / 5) is less than 1, then the value range of a is?
1=loga(a)
loga(2/5)
If the equation x2 + ky2 = 2 represents an ellipse with focus on the y-axis, then the value range of the real number k is ()
A. （0，+∞）B. （0，2）C. （1，+∞）D. （0，1）
∵ equation x2 + ky2 = 2, that is, X22 + y22k = 1 means that the ellipse with focus on the y-axis ∵ 2K > 2, so 0 < K < 1, so D is selected
Logarithm of 2 / 3
Logarithm of 2 / 3
The equation x ^ 2 / |a | - 1 + y ^ 2 / A + 3 = 1 represents the ellipse with the focus on the X axis, then the value range of the real number a is
|A | - 1 > A + 3 > 0, solution - 2 > a > - 3
Let f (x) = loga (2x + 1) satisfy f (x) > 0 in the interval (- 12, 0). (1) find the value range of real number a; (2) find the monotone interval of function f (x); (3) solve the inequality f (x) > 1
(1) Because x ∈ (- 12, 0), so 0 ＜ 2x + 1 ＜ 1, and f (x) ＞ 0, so 0 ＜ a ＜ 1. (2) because 0 ＜ a ＜ 1, so the monotone decreasing interval of the function is (- 12, + ∞); (3) f (x) = loga (2x + 1) ＞ 1, and because 0 ＜ a ＜ 1, so 0 ＜ 2x + 1 ＜ a, the solution is: - 12 ＜ x ＜ a − 12, so the solution set of the original inequality is: {x |: - 12 ＜ x ＜ a − 12}
If the equation x ^ 2 / (m ^ 2 + 1) + y ^ 2 / (3-m) = 1 represents an ellipse with focus on the Y axis, then the value range of the real number m is?
-2～1
Find the inequality about X: the logarithm of the square-3 of log with a as the base is greater than the logarithm of log with a as the base 2x
A > = 1 and 02x
X ^ 2-2x-3 > 0
(x+1)(x-3)>0
x> 3 or X
Let's know that O is the origin of the coordinate, a is the inverse scale function, y = minus 3 / x, a point on the image, AB is perpendicular to the Y axis, and the perpendicular foot is B, then what is the area of the triangle AOB
Answer: 3 / 2
Analysis: let a (a, b), then the area of triangle AOB is equal to the absolute value of a * b * 1 / 2
Because a is on hyperbola, a * b = - 3
The area of triangle is 3 / 2
The logarithm of solution inequality log with (2x-1) as the base (x ^ 2-x-5) > 0 (sharp,
Such as the title
log2x-1(x^2-x-5)>0
log2x-1(x^2-x-5)>log2x-1(1)
When 0
RELATED INFORMATIONS
- 1. When the inverse scale function y = K / X (k > 0) is known, the image passes through point a (2, m), where ab ⊥ X axis passes through point B, and the area of △ AOB is 3 [1] Find the direct [2] point C [x, y] of K and m on the image of inverse scale function y = K / x, find the value range of function value y when 1 ≤ x ≤ 3, take d [1,0] on the y-axis to connect BD, make a square with BD as the edge in the first quadrant, bdef asks whether e, f fall on the image of y = K / X
- 2. Given that the line L passes through the point P (2,1), and intersects with the positive half axis of X axis and Y axis at two points a and B respectively, and O is the coordinate origin, then the minimum area of triangle OAB is? Why is 2 / A + 1 / b greater than or equal to 2 √ 2 / AB
- 3. What is the area of the triangle ABC when the line y = four-thirds + 4 intersects with X at a, intersects with y at B, and O is the origin There are a few more. Give me 20 more ① What is the image line that the image of a linear function y = - 2x + 3 does not pass through ② It is known that the image of positive scale function y = K1 x intersects with the image of y = k2-9 at point P (3, - 6) If the first-order function y = k2x-9 intersects the x-axis at point a, the coordinates of a can be obtained,
- 4. As shown in the figure, a and B are any two points symmetrical about the origin on the graph of function y = 1 x, AC ∥ Y axis, BC ⊥ Y axis, then the area s of △ ABC=______ .
- 5. If we know that a and B are on the image of quadratic function y = AX2, the abscissa of these two points are - 2, 1 respectively, and the triangle AOB is a right triangle (o is the origin of coordinates), we can find the value of A
- 6. If the distance between the vertex A and the positive half axis of the graph a and B is the maximum distance on the positive half axis of the graph A. 32a−12aB. 32a+12aC. 62a−12aD. 62a+12a
- 7. Given that the right angle vertex a of RT triangle ABC moves on the straight line ρ cos θ = 9, C is the origin, and the angle ACB is 30 °, the polar coordinate equation of the trajectory of vertex B is obtained
- 8. Given that the length of the line segment l passing through point P (2,3) and cut by two parallel lines 3x + 4y-7 = 0 and 3x + 4Y + 8 = 0 is d 1), find the minimum value of D, 2) when the line L and 2) When the line L is parallel to the X axis, try to find the value of D
- 9. It is known that the integer part of 5 - √ 5 is a, and the decimal part is B. find a & # 178; + AB + B & # 178;
- 10. 6274 times the 5th power of (1 + x) equals 15031. Find out how much x equals! Keep 4 decimal places 1090 times the 5th power of (1 + x) is equal to = 3276. Find out how much x is equal to! Keep 4 decimal places 4233 times the 5th power of (1 + x) is equal to = 10496. Find out how much x is equal to! Keep 4 decimal places 3093 times the 5th power of (1 + x) is equal to = 5503. Find out how much x is equal to! Keep 4 decimal places 11597 times the 5th power of (1 + x) is equal to = 34089. Find out how much x is equal to! Keep 4 decimal places If there is a way, it would be better! 3Q
- 11. If the line y = K (x + 1) + 1 and the ellipse x2 / 5 + Y2 / M = 1 always have a common point, and the focus of the ellipse is on the x-axis, then the value range of M is
- 12. 1. It is known that for K ∈ R, the line y = KX + 1 and the ellipse (x ^ 2) / 5 + (y ^ 2) / M = 1 always have a common point, then the value range of the real number m is_______ . How can I figure it out wrong A (0,1) B (0,5) C [1,5] ∪ (5, positive infinity) d [1,5] I figured it out. It's wrong for me
- 13. How to calculate the distance from parabolic focus to hyperbolic asymptote?
- 14. It is known that the left and right focal points of hyperbola are F1F2, the eccentricity is following sign 2 and passing through point (4, - following sign 10) 1 to solve the equation of hyperbola
- 15. It is known that the asymptote equation of hyperbola C is y = ± 3x, and the distance from the right focus f (C, 0) to the asymptote is 3. (1) find the equation of hyperbola C; (2) make a straight line with slope k through F, l intersects the hyperbola at two points a and B, and the central perpendicular of line AB intersects the X axis at D
- 16. A focal coordinate is F1 (0, - 13), and the absolute value of the difference between a point P and two focal distances on the hyperbola is 24
- 17. To solve the hyperbolic equation, it has the same focus as the ellipse x ^ 2 + 4Y ^ 2 = 64, and the absolute value of the distance difference between the point on the hyperbola and the two focuses is 1?
- 18. It is known that the hyperbola C1 and the ellipse C2: x ^ 2 / 49 + y ^ 2 / 36 = 1 have a common focus, and the hyperbola C1 passes through M (3 √ 3,2 √ 2), Then the equation of hyperbola C1 is
- 19. Find the elliptic equation of two foci passing through point a (3,4) and hyperbola X & # 178 / 6-y & # 178 / 3 = 1 Wrong. It's not an ellipse, it's a circle
- 20. Taking the vertex of ellipse X & # 178 / / A & # 178; + Y & # 178 / / B & # 178; = 1 (a > b > 0) as the focus, the hyperbolic equation with the focus as the vertex is?