Given the function y = 3 / (4x + 8), then the value of the independent variable x is?

Given the function y = 3 / (4x + 8), then the value of the independent variable x is?

X is not equal to - 2
X is not equal to 2
The elliptic equation with the same focus as the ellipse 9x2 + 4y2 = 36 and the minor axis length of 45 is______ .
Ellipse 9x2 + 4y2 = 36, C = 5, the focus of ellipse is the same as that of ellipse 9x2 + 4y2 = 36, the half focal length of ellipse is C = 5, that is, A2-B2 = 5, the length of minor axis is 45, B = 25, a = 5, the standard equation of ellipse is y225 + x220 = 1, so the answer is y225 + x220 = 1
In the function y = 4x + 3x / x + 3, the value range of the independent variable x is --- steps to be taken
y=4x+(x+3)/3x
According to the fact that the denominator cannot be 0, we get the following result:
3x≠0
x≠0
So:
The value range of independent variable x is x ≠ 0
Given that the center of the ellipse is at the origin, its left focus F1 coincides with the focus of the parabola y square = - 4x, M is an intersection of the ellipse and the parabola, and the absolute value of f1m = 3 √ 2-3, the elliptic equation is solved
Given that the center of the ellipse is at the origin, its left focus F1 coincides with the focus of the parabola y square = - 4x, M is an intersection of the ellipse and the parabola, and the absolute value of f1m = 3 √ 2-3, the elliptic equation is solved
Analysis: let the elliptic equation be x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0)
∵ its left focus F1 coincides with the focus of parabola y = - 4x
∴F1(-1,0)
∵ m is an intersection of ellipse and parabola, | f1m | = 3 √ 2-3
Let m (x0, Y0)
(x0+1)^2+y0^2=(3√2-3)^2==>(x0-1)^2=(3√2-3)^2
Ψ x0-1 = 3 √ 2-3 = = > x0 = 3 √ 2-2 (rounding), x0 = 4-3 √ 2
∵a^2-b^2=1==>a^2=b^2+1
Substitute the ellipse to get x ^ 2 / (b ^ 2 + 1) - 4x / b ^ 2 = 1 = = > b ^ 2x ^ 2-4b ^ 2x-4x = B ^ 2 (b ^ 2 + 1)
==>b^4-(x^2-4x-1)b^2+4x=0
Ψ B ^ 2 = {(x ^ 2-4x-1) - √ [(x ^ 2-4x-1) ^ 2-16x]} / 2 (rounding) B ^ 2 = {(x ^ 2-4x-1) + √ [(x ^ 2-4x-1) ^ 2-16x]} / 2
Substituting x0 into B ^ 2 = 1
The elliptic equation is x ^ 2 / 2 + y ^ 2 = 1
Given that the center of the ellipse is at the origin, its left focus F1 coincides with the focus of the parabola y ^ 2 = - 4x, M is an intersection of the ellipse and the parabola, and | f1m | = 3 √ 2-3, the elliptic equation is solved
Is that the score?
simple
It is known that the value range of the independent variable X of the function y = root sign x-4x + 4m X-2 is
A. m greater than 1 B. m less than 1 C. m greater than or equal to 1 d m less than or equal to 1 the answer is a does not understand
X-4x + 4m > 0
△=16-16m1
Square - 4x + 4m > 0
△=16-16m1
The focal point of the ellipse is on the coordinate axis, the midpoint of the two focal points is the origin, and the ellipse passes (Geng 6,1), (- Geng 3, - Geng 2), so as to find the equation of the ellipse
6/a^2+1/b^2=1,(1)
3/a^2+2/b^2=1,(2)
(1)-(2)*2,
b^2=3,
a^2=9,
The elliptic equation is: x ^ 2 / 9 + y ^ 2 / 3 = 1
Two points in, there are 6 / A ^ 2 + 1 / b ^ 2 = 1, 3 / A ^ 2 + 2 / b ^ 2 = 1
The solution is a ^ 2 = 9, B ^ 2 = 3
The equation is x ^ 2 / 9 + y ^ 2 / 3 = 1
If the line y = - 14x + B is the tangent of the function f (x) = 1X, then the real number B=______ .
Since the derivative y ′ = - 1x2 of function f (x) = LX, if the line y = - 14x + B and function f (x) = LX are tangent to P (m, n), then − 14 = - 1m2n = 1m & nbsp; n = - 14 · m + B & nbsp; the solution is m = 2, n = 12, B = 1 or M = - 2, n = - 12, B = - 1. In conclusion, B = ± 1, so the answer is: 1 or - 1
It is known that the ellipse with the center at the origin and the focus on the coordinate axis passes through M (1,4 √ 2 / 3), n (- 3 √ 2 / 2, √ 2)
(1) Find the eccentricity of ellipse;
(2) Is there a point P (x, y) to a fixed point a (a, 0) (where 0) on the ellipse
(1) Let the elliptic equation be MX ^ 2 + NY ^ 2 = 1, substituting the coordinates of known points into m + 32n / 9 = 1; 9m / 2 + 2n = 1, the solution is m = 1 / 9, n = 1 / 4, so the elliptic equation is x ^ 2 / 9 + y ^ 2 / 4 = 1, because a ^ 2 = 9, B ^ 2 = 4, so C ^ 2 = a ^ 2-B ^ 2 = 5, then the eccentricity e = C / a = √ 5 / 3. (2) let P (3C
If the tangent equation of function f (x) = ax ^ 3 + 2x + 1 (a ≠ 0) at x = 1 is x + y-m = 0, then the real number a is equal to
The derivative of F (x) is f '(x) = 3ax ^ 2 + 2, and the derivative at x = 1 (tangent slope) = 3A + 2. Later, the tangent equation tells you that 3A + 2 = - 1, so a = - 1
It is known that the midpoint of the ellipse is at the origin, the focus is on the coordinate axis, and the length of the major axis is three times that of the minor axis. When the ellipse passes through the point m (0, - 3), the equation of the ellipse is solved
I'm in a hurry
The length of the long axis is three times that of the short axis
2a=3*2b
a=3b
a^2=9b^2
If the focus is on the x-axis
x^2/a^2+y^2/b^2=1
Over M (0, - 3)
So B = 3
a=3b=9
x^2/81+y^2/3=1
If the focus is on the y-axis
y^2/a^2+x^2/b^2=1
Over M (0, - 3)
So a = 3
b=a/3=1
a=3b=9
x^2+y^2/3=1
So x ^ 2 / 81 + y ^ 2 / 3 = 1 and x ^ 2 + y ^ 2 / 3 = 1
The intersection of circle and axis is either the value of a or the value of B. X^2+9Y^2=9 9X^2+Y^2=81
Make an oral calculation Let x ^ 2 / M + y ^ 2 / N = 1 (M > 0, n > 0)
Then M = 9N or n = 9m m points are brought into the equation N=9
M = 81 or M = 1
The equation is
X ^ 2 / 81 + y ^ 2 / 9 = 1 or x ^ 2 + y ^ 2 / 9 = 1
PS: M = a ^ 2 n = B ^ 2 or M = B ^ 2 n = a ^ 2
To avoid chaos, let Mn and ab be OK
On the first floor, B has no square