Conic Sections - 2012 Book Archive

190 Pages·2012·7.42 MB·English

This is “Conic Sections”, chapter 8 from the bookAdvanced Algebra(index.html)(v. 1.0).
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i
Chapter 8
Conic Sections
1788
Chapter 8 Conic Sections
8.1 Distance, Midpoint, and the Parabola
LEARNING OBJECTIVES
1. Apply the distance and midpoint formulas.
2. Graph a parabola using its equation given in standard from.
3. Determine standard form for the equation of a parabola given general
form.
Conic Sections
Aconic section1is a curve obtained from the intersection of a right circular cone
and a plane. The conic sections are the parabola, circle, ellipse, and hyperbola.
The goal is to sketch these graphs on a rectangular coordinate plane.
1.A curve obtained from the
intersection of a right circular
cone and a plane.
1789
Chapter 8 Conic Sections
The Distance and Midpoint Formulas
We begin with a review of thedistance formula2. Given two points(x1, y1) and
(x2, y2) in a rectangular coordinate plane, the distancedbetween them is given
by the distance formula,
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
d = √(x2 − x1)2 + (y2 − y1)2
Furthermore, the point that bisects the line segment formed by these two points is
called themidpoint3and is given by the formula,
x + x y + y
1 2 1 2
,
( 2 2 )
The midpoint is an ordered pair formed by the average of thex-values and the
average of they-values.
2.Given two points(x1, y1)
and(x2, y2), the distanced
between them is given by
d =
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯.
√(x2 − x1)2 + (y2 − y1)2
3.Given two points(x1, y1)
and(x2, y2), the midpoint is
an ordered pair given by
x1+x2 , y1+y2 .
( 2 2 )
8.1 Distance, Midpoint, and the Parabola 1790
Chapter 8 Conic Sections
Example 1
Given(−2,−5)and(−4,−3)calculate the distance and midpoint between
them.
Solution:
In this case, we will use the formulas with the following points:
(x1, y1) (x2, y2)
(−2,−5)(−4,−3)
It is a good practice to include the formula in its general form before
substituting values for the variables; this improves readability and reduces the
probability of making errors.
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
d=√(x2 − x1)2 + (y2 − y1)2
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
2 2
=√[−4 − (−2)] + [−3 − (−5)]
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
2 2
=√(−4 + 2) + (−3 + 5)
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
2 2
= (−2) + (2)
√
⎯⎯⎯⎯⎯⎯⎯⎯
=√4 + 4
⎯⎯
=√8
⎯⎯
=2√2
Next determine the midpoint.
8.1 Distance, Midpoint, and the Parabola 1791
Chapter 8 Conic Sections
x + x y + y −2 + (−4) −5 + (−3)
1 2 1 2
, = ,
( 2 2 ) ( 2 2 )
−6 −8
= ,
( 2 2 )
=(−3, − 4)
Plotting these points on a graph we have,
⎯⎯
Answer: Distance:2√2 units; midpoint:(−3,−4)
8.1 Distance, Midpoint, and the Parabola 1792
Chapter 8 Conic Sections
Example 2
The diameter of a circle is defined by the two points(−1,2)and(1,−2).
Determine the radius of the circle and use it to calculate its area.
Solution:
Find the diameter using the distance formula.
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
d=√(x2 − x1)2 + (y2 − y1)2
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
2 2
=√[1 − (−1)] + (−2 − 2)
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
2 2
= (2) + (−4)
√
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
=√4 + 16
⎯⎯⎯⎯
=√20
⎯⎯
=2√5
Recall that the radius of a circle is one-half of the circle’s diameter. Therefore,
⎯⎯
ifd = 2√5units, then
⎯⎯
d 2√5 ⎯⎯
r = = = √5
2 2
The area of a circle is given by the formulaA = πr2 and we have
8.1 Distance, Midpoint, and the Parabola 1793
Chapter 8 Conic Sections
⎯⎯ 2
A =π √5
( )
=π ⋅ 5
=5π
Area is measured in square units.
⎯⎯
Answer: Radius:√5units; area:5πsquare units
Try this!Given(0,0)and(9,−3)calculate the distance and midpoint between
them.
⎯⎯⎯⎯
Answer: Distance:3√10units; midpoint:(9 , − 3)
2 2
(click to see video)
The Parabola
Aparabola4is the set of points in a plane equidistant from a given line, called the
directrix, and a point not on the line, called the focus. In other words, if given a line
Lthe directrix, and a pointFthe focus, then(x,y)is a point on the parabola if the
shortest distance from it to the focus and from it to the line is equal as pictured
below:
4.The set of points in a plane
equidistant from a given line, The vertex of the parabola is the point where the shortest distance to the directrix
called the directrix, and a point
is at a minimum. In addition, a parabola is formed by the intersection of a cone with
not on the line, called the
an oblique plane that is parallel to the side of the cone:
focus.
8.1 Distance, Midpoint, and the Parabola 1794
Chapter 8 Conic Sections
Recall that the graph of a quadratic function, a polynomial function of degree 2, is
parabolic. We can write the equation of aparabola in general form5or we can
write the equation of aparabola in standard form6:
GeneralForm StandardForm
y = ax2 + bx + c y = a(x − h)2 + k
Herea,b, andcare real numbers,a ≠ 0.Both forms are useful in determining the
general shape of the graph. However, in this section we will focus on obtaining
standard form, which is often calledvertex form7. Given a quadratic function in
standard form, the vertex is(h,k).To see that this is the case, consider graphing
y = (x + 3)2 + 2using transformations.
5.The equation of a parabola
written in the form
y = ax2 + bx + cor y=x2 Basicsquaringfunction.
x = ay2 + by + c, wherea,
y=(x + 3)2 Horizontalshiftleft3units.
b, andcare real numbers and
a ≠ 0. y=(x + 3)2 + 2 Verticalshiftup2units.
6.The equation of a parabola
written in the form
2
y = a(x − h) + kor Use these translations to sketch the graph,
2
x = a(y − k) + h.
7.The equation of a parabola
written in standard form is
often called vertex form. In
this form the vertex is
apparent:(h,k).
8.1 Distance, Midpoint, and the Parabola 1795
Chapter 8 Conic Sections
Here we can see that the vertex is(−3,2).This can be determined directly from
the equation in standard form,
y= a(x − h)2+k
⏐ ⏐
↓ ↓
y=[x − (−3)]2+2
Written in this form we can see that the vertex is(−3,2).However, the equation is
typically not given in standard form. Transforming general form to standard form,
by completing the square, is the main process by which we will sketch all of the
conic sections.
8.1 Distance, Midpoint, and the Parabola 1796