From the point a (- 1,4), make the tangent l of the square of the circle (X-2) + (Y-3) = 1, and find the tangent length From point a (- 1,4) make the tangent l of the square of the circle (X-2) + (Y-3) = 1

From the point a (- 1,4), make the tangent l of the square of the circle (X-2) + (Y-3) = 1, and find the tangent length From point a (- 1,4) make the tangent l of the square of the circle (X-2) + (Y-3) = 1

1: Using Pythagorean theorem
The center coordinates are (2,3) and the radius is 1
The distance from point a (- 1,4) to the center of the circle is
√[(2+1)^2+(3-4)^2]=√10
Then tangent length = √ [(√ 10) ^ 2-1 ^ 2] = 3
2: Geometric method: because the ordinate of point a is equal to the highest point of the circle, the tangent is parallel to the x-axis, and the tangent length is the length between the abscissa of point a and the abscissa of the center of the circle: 2 - (- 1) = 3
If the ellipse x ^ 2 / 16 + y ^ 2 / 9 = 1 and the hyperbola x ^ 2 / (m ^ 2-8) - y ^ 2 / 2m = 1 have the same focus, then the value of M?
∵x^2/16+y^2/9=1
c^2=16-9=7
The hyperbola x ^ 2 / (m ^ 2-8) - y ^ 2 / 2m = 1
(m ^ 2-8) + 2m = 7 and m ^ 2-8 > 0,2m > 0
==>m^2+2m-15=0
==>M = - 5 (rounding off), M = 3
So m = 3
If hyperbola x ^ 2 / M-Y ^ 2 / 2m = 1 and ellipse x ^ 2 / 5 + y ^ 2 / 30 = 1 have the same focus, then M=____
Ellipse A & # 178; = 30, B & # 178; = 5; so C & # 178; = 25; the focus is on the Y axis, so the lower focus F1 (0, - 5); the upper focus F1 (0,5); the lower focus F1 (0,5); the upper focus F1 (0,5); the lower focus F1 (0,5); the lower focus F1 (0,5); the upper focus F1 (0,5); the lower focus F1 (0,5); the upper;
If the hyperbola and the ellipse have the same focus, then the real axis of the hyperbola should be the Y axis and the imaginary axis should be the X axis
Logarithm of negative log with base 3 and base 5? (calculation process)
-log(3)(5)
=log(3)(5^-1)
=log(3)(1/5)
Formula:
alog(N)(M)=log(N)(M^a)
1. Bottom change: original = - (lg5 / Lg3)
2. Look up the logarithm table: lg5 = 0.699, Lg3 = 0.4771.
-0.699/0.4771=-1.465
Can I draw two different inverse scale function images in the same rectangular coordinate system, or do I have to draw two graphs
It can't be in the same rectangular coordinate system. In y = K / x, y and X have the same sign, in one or three quadrants; in y = - K / x, y and X have different sign, in two or four quadrants
Is log a value? What does it represent? Logx does not equal X in value
2. How to deduce the inference formula of the bottom changing formula
Log is the sign of logarithm, logx is the logarithm function, X is a function, the two are not equal, the bottom exchange formula is a more important formula, in many logarithm calculation are used, but also the focus of high school mathematics. Log (a) (b) represents the logarithm of B with a as the base. The so-called bottom exchange formula is log (a) (b) = log (n) (b)
Given the intersection point P (3, - 6) between the image of positive scale function y = K & # 185; X and the image of primary function y = K & # 178; X-9, we can find the value of K & # 185;, K & # 178
Substituting (3, - 6) into the two analytic expressions, we get: - 6 = 3k1, - 6 = 3k2-9
The solution is: K1 = - 2, K2 = 1
3k1 = - 6, K1 = - 2: 3k2-9 = - 6, K2 = 1
If log is based on 7 [log is based on 3 (log is based on 2, x)] = 0, then x ^ 1 / 2 equals 0
Analysis,
log(7)[log(3){log(2)x}]=0=log(7)1
Therefore,
log(3)[log(2)x]=1=log(3)3
Therefore,
log(2)x=3
That is, x = 8
x^(1/2)=2√2.
It is known that the primary function y = (4-K) x + K & # 178; - 9 is a positive proportional function passing through two or four quadrants, and the value of K is obtained
∵ a straight line passes through two or four quadrants
∴4-k4
It is also a positive proportion function
∴k^2-9=0
The solution: k = ± 3
There is no solution for K
-2 is less than or equal to log (1 / x) (10), log (1 / x) (10) is greater than - 1 / 2
Why is this equation equal to
Log (1 / x) (- 2) ≤ log (1 / x) (10) greater than log (1 / x) (- 1 / 2)
The meaning of the building owner is to ask: - 2 ≤ log (1 / x) (10), log (1 / x) (10) > - 1 / 2 is equivalent to log (1 / x) (- 2) ≤ log (1 / x) (10), log (1 / x) (10) > log (1 / x) (- 1 / 2) this is wrong! How can log (1 / x) (- 2) be equivalent to - 2? How can log (1 / x) (- 1 / 2) be equivalent to - 1 / 2
For what?