1. 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
4. Another way to see conics, and you canAnother way to see conics, and you can
also try this at home with a Styrofoam cup.also try this at home with a Styrofoam cup.
6. CircleCircle
The ferris wheel isThe ferris wheel is
an example of aan example of a
circle. Thiscircle. This
picture shows thepicture shows the
first ferris wheelfirst ferris wheel
created in 1893created in 1893
by George W.by George W.
Ferris. The wheelFerris. The wheel
had a diameter ofhad a diameter of
250 feet and250 feet and
circumference ofcircumference of
825 feet.825 feet.
7. 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 =+
8. 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
P
22
)()( 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
9. 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
4
1
4
1
)
2
1
()
2
1
(
4
1
4
1
4
1
4
1
Sehingga dari persamaan diatas dapat diperoleh titik pusat dan jari2 lingkaran
)
2
1
,
2
1
( BA −− dan CBAr −+= 22
4
1
4
1
10. 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
11. 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
2
2
2
1
1
1
mrmxy
mrmxy
+−=
++=
12. 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
+−−=−
++−=−
13. 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)
2
1
2
1 )()( kyhxPQ −+−=
Gradien
hx
ky
PQ
−
−
=
1
1
Karena garis singgung saling tegak lurus dengan PQ maka gradien garis
singgung tersebut adalah
ky
hx
−
−
−
1
1
Sehingga persamaan garis singgung yang dicari adalah Ax +By = C,
C = konstan
cykyxhx =−+− )()( 11
14. 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 =−+− 2
1
2
1 )()(
2
11 ))(())(( rkykyhxhx =−−+−−
Jika lingkaran berpusat di O(0,0) maka persamaan garis singgung
lingkaran di titik Q(x1,y1)
2
11 ryyxx =+
15. 1. Sketch the circle (x − 2)2
+ (y − 3)2
= 16
answer
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.
16. 2. Find the points of intersection of the circle x2
+ y2
− x − 3y = 0 with the line
y = x − 1.
Answer
We solve the 2 equations simultaneously by substituting the expression
y = x -1
into 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).
17.
18. L
P
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
19. Conic Sections - Parabola
The intersection of a
plane with one nappe
of the cone is a
parabola.
20. 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.)
21. 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.
22. Conics used in real life.
• The parabola is in the McDonalds sign.
24. Paraboloid Revolution
They are commonly
used today in satellite
technology as well as
lighting in motor vehicle
headlights and
flashlights.
25.
26. 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”.
27. Conic Sections - Parabola
The line is called the directrix and the point is called
the focus.
Focus
Directrix
28. 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
29. 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
30. Each of the colour-coded line segments is the same length in
this spider-like graph:
31. 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
= 4py
In more familiar form, with "y = " on the left, we can write this as:
PARABOLA DENGAN SUMBU VERTIKAL
32. PARABOLA DENGAN SUMBU HORISONTAL
Dengan cara yang sama pada parabola dengan sumbu vertikal diperoleh
persaman parabola dengan sumbu horisontal
y2
= 4px
33. Shifting the Vertex of a Parabola from the Origin
This 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)
34. If the axis of a parabola is horizontal, and the vertex is at (h, k),
the equation becomes
(y − k)2
= 4p(x − h)
35. CONTOH SOAL
1. Sketch the parabola
Find the focal length and indicate the focus and the directrix on your graph.
ANSWER
The 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.
36.
37. 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
= 4px
Since p = -2 (from the question), we
can directly write the equation of the
parabola:
y2
= -8x
38. 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.
39. 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
= 14y
5. 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?
