IRISAN KERUCUT

Sebuah kerucut tegak jika dipotong dengan berbagai bidang yang mempunyai sudut berbeda beda terhadap sumbu simetri akan membentuk kurva antara lain lingkaran, ellips, parabola, dan hiperbola

Next

Parabola Circle Ellipse Hyperbola

6

4

2

-2

-4

-6

-10 -5 5 10

PreviousMain MenuEnd

Conic sectionsConic sections Shown together.Shown together.

Another way to see conics, and you can also Another way to see conics, and you can also try this at home with a Styrofoam cup.try this at home with a Styrofoam cup.

CircleCircle

©National Science Foundation

CircleCircle

The ferris wheel is an The ferris wheel is an example of a circle. example of a circle. This picture shows This picture shows the first ferris wheel the first ferris wheel created in 1893 by created in 1893 by George W. Ferris. George W. Ferris. The wheel had a The wheel had a diameter of 250 feet diameter of 250 feet and circumference of and circumference of 825 feet.825 feet.

LINGKARAN

Lingkaran adalah himpunan titik-titik pada bidang datar yang jaraknya sama panjang dari suatu titik tertentu. Titik tertentu tersebut dinamakan titik pusat dan jarak yang sama tersebut disebut jari-jari lingkaran.

T(x,y)

Lingkaran dengan titik pusat O(0,0) dan mempunyai jari-jari r. titik T(x,y) terletak pada lingkaran maka jarak titik T dan titik O adalah :

O(0,0)

22 YXOT

OT = jari-jari lingkaran = r

Maka diperoleh persamaan lingkaran

rYX 22

222 rYX

T(x,y) Lingkaran dengan titik pusat P(h,k) dan jari-jari r. Jika titik T (x,y) adalah sebarang titik pada lingkaran maka

TP adalah jari-jari lingkaran, maka diperoleh hubungan

P22 )()( kyhxTP

rkyhx 22 )()(

Sehingga persamaan lingkaran yang berpusat di titik P(h,k) dengan jari-jari r adalah

(x − h)2 + (y − k)2 = r2

Persamaan lingkaran (x − h)2 + (y − k)2 = r2.

Bila ruas kiri diuraikan maka diperoleh

x2-2hx+h2+y2-2ky+k2-r2=0

x2+y2-2hx-2ky+h2+k2-r2=0

Atau ditulis dalam bentuk

x2+y2+Ax+By+C=0

Persamaan di atas merupakan bentuk umum persamaan lingkaran. Dari bentuk umum persamaan lingkaran tersebut dapat kita tentukan koordinat titik pusat dan jari-jarinya dengan mengubah persamaan tersebut menjadi

x2+y2+Ax+By = - C

CBAByAx

CBABByyAAxx

2222

222222

41

41)

21()

21(

41

41

41

41

Sehingga dari persamaan diatas dapat diperoleh titik pusat dan jari2 lingkaran

)21,

21( BA dan CBAr 22

41

41

PERSAMAAN GARIS SINGGUNG PADA LINGKARAN

y=mx+n

Sebuah garis lurus dengan persaman y=mx+n sedangkan persamaan lingkaran x2+y2=r2

Garis singgung yang dicari harus sejajar dengan garis y=mx+n

Kita misalkan persamaan garis singgung yang dicari y=mx+k. karena garis L menyinggung lingkaran maka ada sebuah titik yang koordinatnya memenuhi persamaan garis maupun persamaan lingkaran sehingga diperoleh

L

02)1(

02

)(

2222

22222

222

rkmkxxm

rkmkxxmx

rkmxx

y=mx+k

Karena garis singgung pada lingkaran hanya mempunyai satu titik persekutuan maka persamaan kuadrat hanya mempunyai satu harga x, syaratnya diskriminan dari persamaan tersebut harus sama dengan nol

042 ACBD

044444

0)(44

0))(1(44

22222222

22222222

22222

kmkmrkkm

kmkmrkkm

rkmkm

2

222

2222

1

0)1(

0)(4

mrk

mrk

rmrk

Sehingga persamaan garis singgungnya adalah

22

21

1

1

mrmxy

mrmxy

y=mx+n

P

Jika lingkaran tersebut mempunyai titik pusat P(h,k) maka persamaan garis singgung yang sejajar dengan garis y=mx+n adalah

2

2

1)(

1)(

mrhxmky

mrhxmky

P

Q(x1,y1) Pada sebuah lingkaran mempunyai persamaan (x-h)2+(y-k)2=r2 akan dicari persamaan garis singgung di titik Q(x1,y1)

21

21 )()( kyhxPQ

Gradien hxkyPQ

1

1

Karena garis singgung saling tegak lurus dengan PQ maka gradien garis singgung tersebut adalah

kyhx

1

1

Sehingga persamaan garis singgung yang dicari adalah Ax +By = C,

C = konstan cykyxhx )()( 11

Karena titik Q(x1,y1) terletak pada garis singgung lingkaran, maka

ckykyhxhx ))(())(( 1111

atau

Jadi persamaan garis singgung lingkaran dengan pusat P(h,k) adalah

ckyhx 21

21 )()(

211 ))(())(( rkykyhxhx

Jika lingkaran berpusat di O(0,0) maka persamaan garis singgung lingkaran di titik Q(x1,y1)

211 ryyxx

1. Sketch the circle (x − 2)2 + (y − 3)2 = 16answer

The equation is in the form (x − h)2 + (y − k)2 = r2, so we have a circle with centre at (2, 3) and the radius is r = √16 = 4.

2. Find the points of intersection of the circle x2 + y2 − x − 3y = 0 with the line

y = x − 1.

AnswerWe solve the 2 equations simultaneously by substituting the expressiony = x -1into the expression we have

So we see that the solutions for x are x = 1 or x = 2. This gives the corresponding y-vales of y = 0 and y = 1. So the points of intersection are at: (1, 0) and (2, 1).

LP

F

Pada sebuah bidang terdapat garis L (garis arah) dan sebuah titik focus diluar garis L. Himpunan titik-titik P yang perbandingan antara PF dengan PL memenuhi hubungan

PF = e PL

e= keeksentrikan/ eksentrisitas numerik

Apabila

0<e<1 maka kurva berbentuk ellips

e = 1 kurva berbentuk parabola

e > 1 kurva berbentuk hiperbola

Untuk setiap kasus, kurva-kurva tersebut simetri terhadap garis yang melalui fokus dan tegak lurus garis arah yang disebut directrix. Titik potong antara sumbu dengan kurva disebut puncak

Conic Sections - Parabola

The intersection of a plane with one nappe of the cone is a parabola.

Arch Bridges − Almost Parabolic

The Gladesville Bridge in Sydney, Australia was the longest single span concrete arched bridge in the world when it was constructed in 1964. The shape of the arch is almost parabolic, as you can see in this image with a superimposed graph of y = −x2/4p (The negative means the legs of the parabola face downwards.)

ParabolaParabola

• I found the St. Louis Arch to be an example of a parabola. Standing 630 feet above the Mississippi River, the Arch is America’s tallest monument.

Conics used in real life.

• The parabola is in the McDonalds sign.

Parabolas

© Art Mayoff © Long Island Fountain Company

Paraboloid Revolution

They are commonly used today in satellite technology as well as lighting in motor vehicle headlights and flashlights.

Conic Sections - Parabola

The parabola has the characteristic shape shown above. A parabola is defined to be the “set of points the same distance from a point and a line”.

Conic Sections - Parabola

The line is called the directrix and the point is called the focus.

Focus

Directrix

Conic Sections - Parabola

The line perpendicular to the directrix passing through the focus is the axis of symmetry. The vertex is the point of intersection of the axis of symmetry with the parabola.

Focus

Directrix

Axis of Symmetry

Vertex

Conic Sections - Parabola

The definition of the parabola is the set of points the same distance from the focus and directrix. Therefore, d1 = d2 for any point (x, y) on the parabola.

Focus

Directrix

d1

d2

Each of the colour-coded line segments is the same length in this spider-like graph:

Adding to our diagram from above, we see that the distance d = y + p. Now, using the distance formula on the general points (0, p) and (x, y), and equating it to our value d = y + p, we have

Squaring both sides gives:(x − 0)2 + (y − p)2 = (y + p)2

Simplifying gives us the formula for a parabola:x2 = 4pyIn more familiar form, with "y = " on the left, we can write this as:

PARABOLA DENGAN SUMBU VERTIKAL

PARABOLA DENGAN SUMBU HORISONTAL

Dengan cara yang sama pada parabola dengan sumbu vertikal diperoleh persaman parabola dengan sumbu horisontal

y2 = 4px

Shifting the Vertex of a Parabola from the OriginThis is a similar concept to the case when we shifted the centre of a circle from the origin.To shift the vertex of a parabola from (0, 0) to (h, k), each x in the equation becomes (x − h) and each y becomes (y − k).So if the axis of a parabola is vertical, and the vertex is at (h, k), we have (x − h)2 = 4p(y − k)

If the axis of a parabola is horizontal, and the vertex is at (h, k), the equation becomes (y − k)2 = 4p(x − h)

CONTOH SOAL1. Sketch the parabola Find the focal length and indicate the focus and the directrix on your graph. ANSWERThe focal length is found by equating the general expression for y

and our particular example:

So we have:

This gives p = 0.5.So the focus will be at (0, 0.5) and the directrix is the line y = -0.5.

2. Sketch the curve and find the equation of the parabola with focus (-2,0) and directrix x = 2.

Answer

In this case, we have the following graph

After sketching, we can see that the equation required is in the following form, since we have a horizontal axis:y2 = 4pxSince p = -2 (from the question), we can directly write the equation of the parabola:y2 = -8x

1. Find the distance between the points (3, -4) and (5, 7).

2. Find the slope of the line joining the points (-4, -1) and (2, -5).

3. What is the distance between (-1, 3) and (-8, -4)? A line passes through (-3,

9) and (4, 4). Another line passes through (9, -1) and (4, -8). Are the lines

parallel or perpendicular?

4. Find k if the distance between (k,0) and (0, 2k) is 10 units.

9. Find the equation of the line that passes through (-2, 1) with slope of -3.

10.What is the equation of the line perpendicular to the line joining (4, 2) and

(3, -5) and passing through (4, 2)?

11.Draw the line 2x + 3y + 12 = 0.

12. If 4x − ky = 6 and 6x + 3y + 2 = 0 are perpendicular, what is the value of k?

13.Find the perpendicular distance from the point (5, 6) to the line -2x + 3y + 4

= 0, using the formula we just found.

1. Find the equation of the circle with centre (3/2, -2) and radius 5/2.

2. Determine the centre and radius and then sketch the circle: 3x2 + 3y2 − 12x + 4 = 0

3. Find the points of intersection of the circle x2 + y2 − x − 3y = 0 with the line y = x − 1.

4. Sketch x2 = 14y5. We found above that the equation of the parabola with

vertex (h, k) and axis parallel to the y-axis is (x − h)2 = 4p(y − k). Sketch the parabola for which (h, k) is (-1,2) and p = -3.

6. A parabolic antenna has a cross-section of width 12 m and depth of 2 m. Where should the receiver be placed for best reception?