math1.2
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1.3 : Parabolas
If a plane intersects the cone when it is slanted the same as the side of the cone, (formally, when it is parallel to the slant height), the conic section is a parabola. This is shown below:
Parabolas
Parabola: the set of points in a plane that are the same distance from a given point called the focus and a given line called the directrix.
Directrix
The light source is theFocus
The cross section of a headlightis an example of a parabola...
Here are some other applications of the parabola...
Directrix
Focus
d1
d1
d2
d2
d3
d3
Also, notice that the distance from the focus to any point on the parabola is equal to the distance from that point to the directrix...
We can determine the coordinates of the focus, and the equation of the directrix, given the equation of the parabola....
Vertex
Notice that the vertex is located at the midpoint between the focusand the directrix...
Standard Equation of a Parabola: (Vertex at the origin)
Equation Focus Directrix
x2 = 4py (0, p) y = –p
Equation Focus Directrix
y2 = 4px (p, 0) x = –p
(If the x term is squared, the parabola is up or down)
(If the y term is squared, the parabola isleft or right)
Examples: Determine the focus and directrix of the parabola y = 4x2 :Since x is squared, the parabola goes up or
down…
Solve for x2 x2 = 4py
y = 4x2
4 4 x2 = 1/4y
Solve for p 4p = 1/4
p = 1/16
Focus: (0, p) Directrix: y = –p
Focus: (0, 1/16) Directrix: y = –1/16Let’s see what this parabola looks like...
Examples: Determine the focus and directrix of the parabola –3y2 – 12x = 0 :Since y is squared, the parabola goes left or
right…
Solve for y2 y2 = 4px
–3y2 = 12x –3y2 = 12x
–3 –3
y2 = –4x
Solve for p 4p = –4
p = –1
Focus: (p, 0) Directrix: x = –p
Focus: (–1, 0) Directrix: x = 1Let’s see what this parabola looks like...
Examples: Write the standard form of the equation of the parabola with focus at (0, 3) and vertex at the origin.
Since the focus is on the y axis,(and vertex at the origin) the parabola goes up or down…
x2 = 4py
Since p = 3, the standard form of the equation is x2 = 12y
Example: Write the standard form of the equation of the parabola with directrix x = –1 and vertex at the origin.
Since the directrix is parallel to the y axis,(and vertex at the origin) the parabola goes left or right…
y2 = 4px
Since p = 1, the standard form of the equation is y2 = 4x
Equation Parabola at (h,k)
If the vertex of the parabola is at (h,k), the standard equation for the parabola are as summarised below.
Parabola Focus Directrix Shape
(y-k)2=4p(x-h) (h+p,k) x=h-p Opens to the right
(y-k)2=-4p(x-h) (h-p,k) x=h+p Opens to the left
(x-h)2=4p(y-k) (h,k+p) y=k-p Opens upwards
(x-h)2=-4p(y-k) (h,k-p) y=k+p Opens downwards
Example 1
State the vertex, focus and directrix of each of
)3(12)2( 2 xy
Solution
(y−2)2 =12(x−3)
Vertex, V(h, k) = V(3, 2)
P= 3, so focus, F( (h+p, 2) P(3+3, 2) P(6, 2
)
Example 4 Sketch the graph of x2 + 8x + 4y + 12 = 0. Solution Write the equation in standard form.
x2 + 8x + 4y + 12 = 0. x2 + 8x + 16 + 4y + 12 – 16 = 0. (x + 4)2 + 4(y – 1) = 0 (x + 4)2 = – 4(y – 1)
This is a parabolic equation with vertex V(– 4, 1), p = – 1 and F(– 4, 0).
When y = 0, x = – 2 or x = – 6. The sketch of the graph is as follow,
Symmetrical axis
directrix
V(–4, 1)
(–6, 0)F(–4, 0)
(–2, 0)
0
Example 5 Find the equation of a parabola which satisfies the following
conditions, vertex , vertical symmetric axis and the parabola passes through point (3,6)
Solution
Standard equation of the parabola is
Vertical symmetric axis The parabola passes through (3,6)
2,1
1h 2k
241 2 ypx
2
8416
26413 2
p
p
p
The equation of the parabola is:
01582
16812
281
2241
2
2
2
yxx
yxx
yx
yx
Example 6
A necklace hanging between two fixed points A
and B at the same level. The length of the
necklace between the two point is 100 cm. The
mid point of the necklaceis 8 cm below A and B.
Assume that the necklace hangs in the form of
parabolic curve, find the equation of the curve.
Solution:
8cm
50cm 50cm
A B
y
x
kyphx 42
.12.7832
2500
)8(450
)8(4)(2
2
cmp
p
yp
ypx
)8(48.3122 yx