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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?