10.1 Conics and Calculus

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Transcript 10.1 Conics and Calculus

10.1
Parabolas
10.1 Parabolas
A parabola is the set of all points (x,y) that are
equidistant from a fixed line (directrix) and a fixed point
(focus) not on the line.
Focus (h, k + p)
Vertex (h,k)
Directrix y = k - p
Standard Equation of a Parabola
(x - h)2 = 4p(y - k)
Vertical axis
Opens up (p is +) or down (p is -)
(y - k)2 = 4p(x - h)
Horizontal axis
Opens right (p is +) or left (p is -)
p is the distance from the center to the focus
point.
Ex.
Find the vertex, focus, and directrix of the parabola
and sketch its graph. y2 + 4y + 8x - 12 = 0
Now complete the
square.
y2 + 4y = -8x + 12
y2 + 4y + 4 = -8x + 12 + 4
(y + 2)2 = -8x + 16
(y + 2)2 = -8(x - 2)
4p = -8
p = -2
Write down the vertex
and plot it. Then find p.
What does the negative p
mean?
left
Directrix
x=4
V(2,-2)
F(0,-2)
Ex.
Find the standard form of the equation of the
parabola with vertex (2,1) and focus (2,4).
First, plot the two points.
Which equation will we be using? Vert. or Horz. axis
Right, since the axis is vertical, we will be using
(x - h)2 = 4p(y - k)
What is p?
p=3
Now write down the equation.
(x - 2)2 = 12(y - 1)
Ellipses
Ellipses
Center point
(h,k)
Focus point
a
F
V
Major axis
a2 = b2 + c2
b
c
F
a
V
Minor axis
An ellipse is the set of all points (x,y), the sum
of whose distances from two distinct points
(foci) is constant.
Standard Equation of an Ellipse
( x  h) ( y  k )

1
2
2
a
b
2
2
( x  h) ( y  k )

1
2
2
b
a
2
2
Horz. Major axis
Vert. Major axis
(h,k) is the center point.
The foci lie on the major axis, c units from the center.
c is found by
c2 = a2 - b2
Major axis has length 2a and minor axis has length 2b.
Sketch and find the Vertices, Foci, and Center point.
x2 + 4y2 + 6x - 8y + 9 = 0
First, write the equation in standard form.
(x2 + 6x +
) + 4(y2 - 2y +
) = -9
(x2 + 6x + 9) + 4(y2 - 2y + 1) = -9 + 9 + 4
(x + 3)2 + 4(y - 1)2 = 4
C (-3,1)
( x  3)
( y  1)

1
4
1
2
2
V (-1,1) (-5,1)
c2 = a2 - b2
C (-3,1)
c2 = 4 - 1
c 3
Foci are:
(3  3,1) & (3  3,1)
V (-1,1) (-5,1)
Eccentricity e of an ellipse measures the ovalness of the
ellipse. e = c/a
In the last example, what is the eccentricity?
The smaller or closer to 0 that the eccentricity is, the more
the ellipse looks like a circle.
The closer to 1 the eccentricity is, the more elongated it is.
Find the center, vertices, and foci of the ellipse given by
4x2 + y2 - 8x + 4y - 8=0
First, put this equation in standard form.
4(x2 - 2x + 1) + ( y2 + 4y + 4) = 8 + 4 + 4
4(x - 1)2 + (y + 2)2 = 16
x  1
2
4
C( ,
a=
b=
c=
)

y  2

2
16
1
Vertices ( , ) ( , )
Foci ( , ) ( , )
e=
Sketch it.
Hyperbolas
The standard form with center (h,k) is
x  h
2
2

y  k

2
2
1
a
b
2
2
 y  k   x  h  1
2
2
a
b
Note: a is under the positive term. It is not necessarily
true that a is bigger than b.
Let’s take a look at the first hyperbola form.
V(h-a,k)
b
c
b
V(h+a,k)
C(h,k)
F(h-c,k)
F(h+c,k)
a
Note: If c is
the distance
from the center
to F, and all radii
of a circle = ,
then the hyp. of the right triangle is also c.
c is the distance
from the center
to the foci.
Therefore, to find c,
a2 + b2 = c2
Sketch the hyperbola whose equation is 4x2 - y2 = 16.
4x2 - y2 = 16
2
2
x
y

1
4 16
First divide by 16.
Write down a, b, c and the center pt.
a=2
b=4
Note: a is always under
the (+) term.
C(0,0)
Now find c.
c2 5
Let’s sketch the hyperbola.
V (2,0)

F  2 5 ,0
F
V
Now, we need to find
the equations of the
asymptotes.
What are their slopes
and one point that is on both lines?
V

F
4
m    2
2
y  0  2( x  0)
Sketch the graph of
4x2 - 3y2 + 8x +16 = 0
4(x2 + 2x +1 ) - 3y2 = -16 + 4
4(x + 1)2 - 3y2 = -12

y
x  1

1
4
3
2
2
C(
,
a=
b=
c=
e=
Sketch
)
Now, divide by -12 and switch
the x and y terms.
C  1,0 
a2
b 3
c 7
7
e
2
V(
F(
,
,
)(
)(
,
,
V  1,2 

F  1, 7
)
)

F
V
V
Eq. of asymptotes.
2
x  1
y0  
3
F
Classifying a conic from its general equation.
Ax2 + Cy2 + Dx + Ey + F = 0
If:
A=C  0
AC = 0 , A = 0 or C = 0, but not both
AC > 0, A  C
AC < 0
Both A and C = 0