Post on 05-Jan-2016
GEOMETRI ANALITIK
PENDIDIKAN MATEMATIKAUNIVERSITAS NEGERI SEMARANG
2013/2014
KELOMPOK 6
YUKEVANNY APRILA P (4101412034) HERLINA ULFA NINGRUM
(4101412088) DYAN MAYLIA (4101412146) RATNA KARTIKASARI (4101412198)
15. For each equation of acircle :a) Write the equation in standard formb) Specify the coordinates of the centerc) Specify the radiusd) Sketch the graph The equation is x2 + y2 + 8x – 6y = 0.
Answer:
25)3()4(
25)96()168(
0...)6(...)8(
068
22
22
22
22
yx
yyxx
yyxx
yxyx
We know that h= -4 , k= 3, and r = 5.
So, the center is at P(-4, 3) and the radius is 5.
From the information, we sketch the graph of circle
X’
X
YY’
O’
O
P(-4,3)
Number 28
16x 2 +25y 2 -64x+100y-236=0
a. Write the equation in standard formb. Specify the coordinates of the center and
verticesc. Graph the curve
Solution :From the equation we know that, it is an equation of ellipse, because A, C
have same signs.
16x2 + 25y2 - 64x + 100y – 236 = 0
16(x2-4x+...) + 25(y2+4y+...) = 236
16(x2-4x+4) + 25(y2+4y+4) =236+ 64 + 100
16(x-2)2 + 25(y+2)2 = 400
So, the coordinates center is P(2,-2)The vertices are Q(-2,-7) and R(-2,-2)To sketch the graph, we write an x’, y’ equation where the origin has been translated to (2,-2)
116
)2(
25
)2( 22
yx
116
'
25
' 22
yx
Graph :
Y’y
x
X’
P(2,-2)
29. For each equation of an elipse or hyperbola:a) Write the equation in standard form,b) Specify the coordinates of the center and vertices,c) Graph the curve. The equation is 01513218169 22 yxyx
Answer:From the equation we know that, it is an equation from a hyperbola.
01513218169 22 yxyx
19
)1(
16
)1(
144)1(16)1(9
169151)12(16)12(9
151...)2(16...)2(9
1513216189
01513218169
22
22
22
22
22
22
yx
yx
yyxx
yyxx
yyxx
yxyx
From the theorem 3.5, we know that:
19
)]1([
16
)]1([
19
)1(
16
)1(
22
22
yx
yx We get h=-1 and k=-1,So the center of the hyperbola is P(-1,-1).
In X’Y’ equation, where the origin has been translated to (-1,-1), so the the equation become:
19
'
16
' 22
yx
The vertices, lie the X’ axis.
Since a2=16 a= ±4 and the vertices are Q(-4,0) and R(4,0) in x’,y’OrThe vertices is A(3,-1) and B(-5,-1) in x,y system.
The graph of hyperbola
P(-1,-1)
X
X’
YY’
O
O’
Q(-4,0)
R(4,0)
30. For each equation of an elipse or hyperbola:a) Write the equation in standard form,b) Specify the coordinates of the center and vertices,c) Graph the curve. The equation is x2 – 9y2 + 54y – 90 = 0
Answer:From the equation we know that, it is an equation of hyperbola, because A, C have different signs.
1)3(9
9)3(9
8190)96(9
90...)6(9
090549
22
22
22
22
22
yx
yx
yyx
yyx
yyx
1)3(9
22
yx
From the theorem 3.5, we know that:
We get h= 0 and k= 3, so the center of the hyperbola is at P(0,3).
In X’Y’ equation, where the origin has been translated to (0, 3), so the the equation become:
1'9
2
2'
yx
The vertices, lie the X’ axis.Since a2=9 a= ±3 and the vertices are Q(-3,0) and R(3,0).
O(0,0)
Y
Y’
X
X’O’
P(3,0)
Number 39
Assuming the graph of each equation exists and is not degenerate,identify its graph using theorem 3.6
Answer: the graph of is an ellips because A≠C and they have the same sign.
014 22 xyx
014 22 xyx
Number 41Write an equation a) in standart form and b) in
general form of parabola with the given vertex, the focus, or directrix
Vertex at (5,6), focus (5,8)
Penyelesaian:
0738102)6(82)5)(
)6(8)5(
)(2)()2
2
yxx
yxb
yx
kyphxa
50. Find a standard equation for a hyperbola satisfying the given condition:Center at (4,6), one vertex at (2,6), one focus at (7,6).
Answer:From the center at (4,6) we know that h=4 and k=6 and we translated the hyperbola to (4,6).x’ = x-h x’= 4-4 =0y’ = y-k y’ = 6-6 =0So, in X’Y’ system the center is O’(0,0).
The vertex in X’Y’ system is:x’ = x-h x’= 2-4 = -2y’ = y-k y’ = 6-6 = 0So, in X’Y’ system the verteces are Q(-2,0) and R(2,0).
The focus in X’Y’ system is:x’ = x-h x’ = 7-4 = 3y’ = y-k y’ = 6-6 = 0So, in X’Y’ system the foci are A(-3,0) and B(3,0).
We have one of the vertex is (2,0), so, a=2. We have one of the focus is (3,0), so c= 3 .Now, we will find the value of c.
c2=a2+b2
b2=c2 – a2
b2 = 9 – 4
b2 = 5
We can find the equation of the hyperbola in X’Y’ system:
15
'
4
'1
'' 22
2
2
2
2
yx
b
y
a
x
Now, we can find the equation of the hyperbola in X,Y system:we know that x’ = x – 4 and y’ = y – 6.
so, the equation is
15
)6(
4
)4( 22
yx
Number 51
Find a standard equation for a hyperbola satisfying the given condition:Vertices at (2,5) and (2,-1), one focus at (2,-4)
The center of hyperbola is midpoint of vertices.
so, the center is P(2,2).and we know that h=2 and k=2.
2
)22(2
12
)( 121
x
x
xxkxx
2
)]1(5[2
11
)( 121
y
y
yykyy
a is distance between center and vertex,
c is distance between center and focus,3
))1(2()22(
)()(
22
212
212
a
a
yyxxa
6
))4(2()22(
)()(
22
212
212
c
c
yyxxc
Therefore,b 2 = c 2 – a2 = 36 – 9 = 27The standard form is :
127
)2(
9
)2( 22
xy
Number 54
Write the equations in theorem 3.3a in general form.
Persamaan tersebut dapat direduksi menjadi
02222222222
22
phkkypxy
phpxkkyy
hxpky )()(
02 FEyDxCy
pkhpyhxx
pkpyhhxx
kyphx
2222222222
)(2)(2
Vertical line of symetry
Persamaan tersebut direduksi menjadi
02 FEyDxx
56. Prove the theorem 3.4If a ellipse has its center at (h,k), then it has an equation
1)()(
1)()(
2
2
2
2
2
2
2
2
b
hx
a
ky
b
ky
a
hx
Proof:From the theorem 2.5, an ellipse has the standard equation
12
2
2
2
b
y
a
x
If only if its center is the origin and itsfoci are in the x axis.
Now, this equation will be translated to (h,k) in X’Y’ system.We know that x = x’+ h x’ = x – h and y = y’ + k y’ = y -kSo, the equation become:
1)()(
2
2
2
2
b
ky
a
hx
57. Prove the theorem 3.5If a hyperbola has its center at (h,k), then it has an equation
1)()(
1)()(
2
2
2
2
2
2
2
2
b
hx
a
ky
b
ky
a
hx
Proof:From the theorem 2.7, a hyperbola has the standard equation
12
2
2
2
b
y
a
x
If only if its center is the origin and itsfoci are in the x axis.
Now, this equation will be translated to (h,k) in X’Y’ system.We know that x = x’+ h x’ = x – h and y = y’ + k y’ = y -kSo, the equation become:
1)()(
2
2
2
2
b
ky
a
hx