Ellipse X & # 178 / M + Y & # 178 / 10 = 1 has the same focus as hyperbola Y & # 178; - X & # 178 / b = 1

Ellipse X & # 178 / M + Y & # 178 / 10 = 1 has the same focus as hyperbola Y & # 178; - X & # 178 / b = 1

The focus is on the y-axis, b > 0,0
If the eccentricity of the ellipse C1: X & sup2; / 4 + Y & sup2; / B & sup2; = 1 is root 3 / 2, the focus of the parabola C2: X & sup2; = 2PY is on the vertex of the ellipse C1
Solving the equation of parabola C2
Because e = C / a = C / 2 = √ 3 / 2, C = √ 3, B = √ [2 & # 178; - (√ 3) &# 178;] = 1
The vertex of the ellipse on the y-axis is (0,1) or (0, - 1)
So the focus of the parabola C2: X & # 178; = 2PY is (0,1) or (0, - 1)
So the equation of parabola C2 is X & # 178; = 4Y or X & # 178; = - 4Y
x^2=4y
x²=4Y HUO x²=-4Y
It is known that the first-order function y = (2-k) x + K & # 178; - 9 is a positive proportional function passing through one or three quadrants, and the value of K is obtained
Because it's a positive scale function, K ^ 2-9 = 0
K = 3 or K = - 3
And because the function passes through one or three quadrants, 2-k > 0 is K
2-k>0,k²-9=0
k=±3,k
Solving logarithmic inequality 1 / (1-lgx) > = 1 / (1 + lgx)
Please write the specific process.
Zero
(0,1 / 10) and up [1,10]
Solution 1 / (1-lgx) > = 1 / (1 + lgx) 1 / (LG (10 / x)) > = 1 / (lg10x)
Discussion by situation
1）0
The properties of positive proportion function y = KX (K ≠ 0)
1. When k > 0, the image passes through______ Quadrant: y increases with the increase of X____ .
2. When k ＜ 0, the image passes through______ Quadrant: y increases with the increase of X____ .
1. When k > 0, the image passes through one, two, four quadrants: y increases with the increase of X
2. When k < 0, the image passes through two, three and four quadrants: y decreases with the increase of X
If log is the logarithm of C with base a + log is the logarithm of C with base B (C ≠ 1), then AB + c-abc is
lgc/lga=-lgc/lgb
c≠1
lgc≠0
So 1 / LGA = - 1 / LGB
lga=-lgb
lga+lgb=0
lgab=0
ab=1
Original formula = 1 + C-C = 1
1. If the positive proportional function y = KX passes through the point (- 1,3), then k =?
2. If the image of a function y = 3x + B passes through a point (1,2), then B =?, the intersection coordinate of the image of the function and the Y axis is?
3. If the image of function Y1 = K1X passes through P (3,4) and is symmetric with the image of function y2 = k2x about y axis, what are their analytical expressions?
4. Find the analytic expression of the function image of (4,0) points
Solution 1: substituting x = - 1, y = 3 into y = KX, we get the following result
3=-k
k=-3
Solution 2: substituting x = 1, y = 2 into y = 3x + B, we get the following result:
2=3×1+b
b=2-3
b=-1
The intersection coordinates of the function and Y axis are (0, b), that is (0, - 1)
Solution 3: substitute x = 3, y = 4 into Y1 = K1X to get 4 = 3k1, K1 = 4 / 3, and then substitute K1 = 4 / 3 into Y1 = K1X to get 4 = 3k1, K1 = 4 / 3
y1=4x/3,
With respect to Y-axis symmetry, then the image of y2 = k2x passes through (3,4) symmetry point (- 3,4) with respect to y-axis
Substitute x = - 3, y = 4 into y2 = k2x to get: 4 = - 3k2, K2 = - 4 / 3, and then substitute K2 = - 4 / 3 into y2 = k2x to get: 4 = - 3k2, K2 = - 4 / 3
y2=-4x/3
Solution 4: because there are countless function images passing through (4,0), it is impossible to determine the analytic expression of the function passing through (4,0)
b=-1
So the function is y = 3x-1. X = 1, y = - 1, that's the intersection coordinates.
Please write down the question clearly, there is ambiguity.
How to compare the logarithm of base 7 with that of base 6?
log7\6-log6\7=ln7/ln6-ln6/ln7
=[(ln7)^2-(ln6)^2]/[ln6 ln7]>0,
So the front one is big
Log7-6 is the logarithm of base 7
If the positive scaling function y = KX passes through points (3, - 4), then K=
(3, - 4) that is, x = 3, y = - 4
So k = Y / x = - 4 / 3
You can get the answer by substituting 3 and - 4
The positive proportional function y = KX passes through the point (3, - 4)
When x = 3, y = - 4
So 3K = - 4
k=-4/3
With 8 as the base, the logarithm of 9 is?
one point zero five six six four one six six seven