假期功课求助(coordinates geometry )

zipp_882000

13)the function f and g are defind by

f(x)=sqrt(2+x),x_>(same or more)  -2
g(x)=sqrt(x^2-4),lxl _>2

find f+g and state the corresponding domian


17)if f(x)=log  (1+x)/1-x)  ,show that
              a

f(x)+f(y)=f {(x+y)/1+xy)

谢谢

[ 本帖最后由 zipp_882000 于 28-12-2006 01:48 PM 编辑 ]

dunwan2tellu

13)the function f and g are defind by

f(x)=sqrt(2+x),x_>(same or more)  -2
g(x)=sqrt(x^2-4),lxl _>2

find f+g and state the corresponding domian

f+g = Sqrt[2+x] + sqrt[x^2-4]
Domain of (f+g) = Domain f  intersect domain g =  x >= 2

17)if f(x)=log  (1+x)/1-x)  ,show that
              a

f(x)+f(y)=f {(x+y)/1+xy)

f(x) = log_a (1+x) - log_a (1-x)
f(y) = log_a (1+y) - log_a (1-y)
f(x) + f(y) = log_a (1+x)(1+y) - log_a (1-x)(1-y)
            = log_a (1+x)(1+y) - log_a(1+xy) - log_a (1-x)(1-y) + log_a(1+xy)
            =log_a (1+x)(1+y)/(1+xy) - log_a (1-x)(1-y)/(1+xy)
            =log_a (x+y+1+xy)/(1+xy) - log_a (1+xy-x-y)/(1+xy)
            =log_a [1 + (x+y)/(1+xy)] - log_a (1 - (x+y)/(1+xy)]
            = f((x+y)/(1+xy))

wjfong

原帖由 dunwan2tellu 于 21-12-2006 09:52 AM 发表

f+g = Sqrt + sqrt
Domain of (f+g) = Domain f  intersect domain g =  x >= 2


f(x) = log_a (1+x) - log_a (1-x)
f(y) = log_a (1+y) - log_a (1-y)
f(x) + f(y) = log_a (1+x)(1+y) - log_a ( ...

这个问题好面熟！应该是很久的pass year哦！

zipp_882000

19)find the equation of circle passes through the ponits A(1,5) B (7.7) a the end of a diameter
i cant get x^2

answer is x^2+y^2-8x-12y+40

220the equation of circle is X2+y2-6x-2y=19
a)find the wheather the point a(4,6)is inside or out side the circle (how to find )


24)the equation of the circle C is x2+y2-18x-6y+45=0

C)determind the coordinate of the contact points B which has the equation of tangen C is y=2x
d)the tangents to C at A and B calculate the coordinates of A and B

28)find the eqaution of 2 straight lines inclined at 30 degree with the x=axis and passes throught a point(-3,-5)

29)two parallel lines I1 passing through (-1,3) and I2  passing through (2,-3) respectively meet the straigtht line y-2x=8 at point P and Q if the distance between Pand Q is qrt5 unit find all possible gradient I1 and I2 .



30) given p(0,6) Q(-3,2)R(2,4) s（h,k) are 4 vertice of parallelogram /.Determine values of h and k .fine the shortest distance from S to the line PR and area of the parallelogram

answer h=5.,K=8 .area=4.95


36)p(ap2,2ap) and Q (aq2,2aq) are 2 points lie on parabola y2=4ax

b) show that the mid-point of all parallel chords with gradient 5 lie on a straight parallel to x-axis


41)the coordinates of the points A and B are (2,1) and (4,5) respectively .Find the equation of locus of a moving point P such that APB is a right- angled triangle

我要怎么知道我的equatuion 是AP=PB ,PA=AB 等等的可能性?

46) two points p(ap2,2ap_ and Q (aq2,2aq) lie on parabola y2=4ax in such a way that POQ is a right angled triangle .where O is origin
a)show that pq=-4
b)If M  is the mid point of PQ ,find the equation of locus M

end

thx

dunwan2tellu

19)find the equation of circle passes through the ponits A(1,5) B (7.7) a the end of a diameter
i cant get x^2

answer is x^2+y^2-8x-12y+40

center = midpoint of AB = (4,6) . Center to A = radius = Sqrt[10]
circle equation : (x-4)^2 + (y-6)^2 = 10

220the equation of circle is X2+y2-6x-2y=19
a)find the wheather the point a(4,6)is inside or out side the circle (how to find )

equation circle : (x-3)^2 + (y-1)^2 = 29
center = (3,1) , radius = Sqrt[29]
distance from point to center = Sqrt[26] < radius ( inside circle)

24)the equation of the circle C is x2+y2-18x-6y+45=0

C)determind the coordinate of the contact points B which has the equation of tangen C is y=2x
d)the tangents to C at A and B calculate the coordinates of A and B

c) substitute y = 2x into equation then use b^2 -4ac = 0
d)A = ??

28)find the eqaution of 2 straight lines inclined at 30 degree with the x=axis and passes throught a point(-3,-5)

30 degree ==> slope = m = + - tan 30
then y = mx + c

29)two parallel lines I1 passing through (-1,3) and I2  passing through (2,-3) respectively meet the straigtht line y-2x=8 at point P and Q if the distance between Pand Q is qrt5 unit find all possible gradient I1 and I2 .

I1 : y-3 = m(x+1) ; I2 : y+3 = m(x-2) ; L : y = 2x + 8
L & I1 : 2x + 5 = m(x+1) => P : x = (m-5)/(2-m) , y = (-6m+6)/(2-m)
L & I2 : 2x+11=m(x-2) ==> Q : x = (11+2m)/(m-2) , y = (12m + 6)/(m-2)
Distance PQ^2 = 5 = (3m+6)^2/(m-2)^2 + (6m+12)^2/(m-2)^2
Solve quadratic ...
  

30) given p(0,6) Q(-3,2)R(2,4) s（h,k) are 4 vertice of parallelogram /.Determine values of h and k .fine the shortest distance from S to the line PR and area of the parallelogram

answer h=5.,K=8 .area=4.95

midpoint PR = midpoint QS 就能得到 h,k
shortest distance : 先找 PR equation , 再用 shortest distance formula i.e
d = |ax+by+c|/Sqrt[a^2+b^2] ...之后想象parallelogram = 2 个triangle , 然后用 base x height x 1/2 = area 来找 triangle area

36)p(ap2,2ap) and Q (aq2,2aq) are 2 points lie on parabola y2=4ax

b) show that the mid-point of all parallel chords with gradient 5 lie on a straight parallel to x-axis

slope PQ = 2a(p-q)/a(p^2-q^2) = 2/(p+q) = 5 ==> p+q = 2/5
midpoint PQ = [ a(p2+q2)/2 , a(p+q) ] = [ a(p2+q2)/2 , 2a/5 ]
y-coordinate = constant = 2a/5 . 所以 parallel to x-axis i.e y = 2a/5

41)the coordinates of the points A and B are (2,1) and (4,5) respectively .Find the equation of locus of a moving point P such that APB is a right- angled triangle

我要怎么知道我的equatuion 是AP=PB ,PA=AB 等等的可能性?

right-angle triangle 是 slope , m1 x m2 = -1 的。大多数是指 P = 90 degree .
所以用 slope PA x slope PB = -1

46) two points p(ap2,2ap_ and Q (aq2,2aq) lie on parabola y2=4ax in such a way that POQ is a right angled triangle .where O is origin
a)show that pq=-4
b)If M  is the mid point of PQ ,find the equation of locus M

a)用 Slope OP x Slope OQ = -1
b)M = midpoint PQ = [ a(p2+q2)/2 , a(p+q) ]

x = a(p2+q2)/2 , y = a(p+q) , pq = -4 ==> p2+q2 = 2x/a

y^2 = a^2 * (p^2+q^2+2pq) = a^2 * (2x/a -8) = 2ax - 8a^2