GEOMETRI ANALITIK

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 2816x 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

22'

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)

YY’

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

yxxyxb

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

by

ax

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(212

)( 121

x

x

xxkxx

2

)]1(5[211

)( 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

aa

yyxxa

6))4(2()22(

)()(22

212

212

cc

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

bhx

aky

bky

ahx

Proof:From the theorem 2.5, an ellipse has the standard equation

12

2

2

2

by

ax

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

bky

ahx

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

bhx

aky

bky

ahx

Proof:From the theorem 2.7, a hyperbola has the standard equation

12

2

2

2

by

ax

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

bky

ahx