BC:x+3y+4=0，

CA:x-5y+ 12=0，

Find the size of (1)∠A;

(2) The straight line equation where the bisector of ∠ a lies;

(3) The linear equation of the height of 3)BC side.

It is known that ABC's trilateral equation is AB: 5x-y- 12 = 0,

BC:x+3y+4=0，

CA:x-5y+ 12=0，

Find the size of (1)∠A;

(2) The straight line equation where the bisector of ∠ a lies;

(3) The linear equation of the height of 3)BC side.

The center of hyperbola is at the origin O, the focus is on the X axis, and the two asymptotes are

, perpendicular to the right focus f.

Straight lines intersect respectively.

Known a and b

、

、

Become arithmetic progression, and then

In the same direction

(1) Find the eccentricity of hyperbola;

(2) Let the line segment length of hyperbola section AB be 4, and find the equation of hyperbola.

Answer (i) when 0.

f′(x)= 1-lnx- 1 =-lnx & gt； 0

So the function f(x) is the increasing function in the interval (0, 1),

(2) When 0

It is also known from (i) that f(x) is continuous at x= 1

When 0

Therefore, when 0

①

The following is proved by mathematical induction:

0< An & ltan+1< 1

②

(1) From 0

A2=f(a 1), and 0 < a1<; A2< 1, that is, when n= 1, inequality ② holds.

(ii) If n=k, inequality ② holds, that is, 0.

Then 0

Therefore, when n=k+ 1, inequality ② also holds.

Combining (i) and (ii) to prove

(3) According to (2), {an} increases item by item, so if there is a positive integer m≤k, so that am≥b, then AK+1>; am≥b

Otherwise, if I

AML Nam≤a 1 lnam & lt； a 1 lnb & lt； 0

③

ak+ 1=ak-aklnak

= AK- 1-AK- 1 lnak- 1-aklnak

……

=a 1-

Amler M.

Understand through ③

amlnam & ltk

(a 1lnb)

So AK+1>; a 1+k|a 1lnb|

≥a 1+(b-a 1)

=b