Unit 2 Conic Sections
By: Areeb Rahim, Vibha Makam,
Jill Vaishnav, Krish Magal
Chapter Overview
1. Apply the midpoint and distance formula
2. Absolute value equations/inequalities
3. Graph absolute value
4. Complete the square
5. Quadratic Equations
6. Parabolas
7. Circles
Apply the midpoint and distance formula
Distance Formula:
Midpoint Formula:
What is the distance between (-5,1)
EXAMPLE: and (-3,2)?
EXAMPLE: Identify △ABC as scalene, equilateral, or scalene.
A(-2,-5) B(0,3) C(6,3)
AB= √(0--2)²+(3--5)² BC= √(6-0)²+(3-3)²
= √4+64
= √36+0
= √68
=6
= 2√17 The triangle is scalene since
all the sides are different
lengths
D= √(-3+5)²+(2-1)²
= √4+1
= √5
EXAMPLE: Use the distance d to find the value of x.
AC= √(6--2)²+(3--5)²
= √64+64
= √128
= 8√2
(0,3), (x,5); d= 2√10
EXAMPLE: Find the midpoint of the segment joining (2,7), (8,2)
2√10= √(x-0)²+(5-3)²
(2√10)²= (√x²+4)²
40= x²+4
36= x²
±6= x
square both sides to get rid of the radical
(2+8/2, 7+2/2)
(10/2, 14/2)
(5,7)
Absolute Value Equations/Inequalities
|x|=
x, when x>0
0, when x=0
-x, when x<0
*when the absolute value equals a
negative number, it is no solution*
EXAMPLE: |x-9|+5=10
isolate the absolute value
|x-9|=5
x-9=5 or x-9=-5
x=14 or x=4
*When solving an inequality, if the absolute value is LESS than the
number, then you make an “and” statement. If the inequality is
GREATER than the absolute value, then you make an “or”
statement*
EXAMPLE: -2|6x-1|+5<3
-2|6x-1|<-2
|6x-1|>1
6x-1>1 or 6x-1<1
6x>2 or 6x<0
x>⅓ or x<0
PRACTICE #1)
Solve the inequality, don’t graph:
-7+2|3x-2|≥9
ANSWER #1) -7+2|3x-3|≥9
2|3x-2|≥16
|3x-2|≥8
3x-2≥8 or 3x-2≤-8
3x≥10 or 3x≤-6
x≥10/3 or x≤-2
PRACTICE #2) Solve and graph the inequality:
3|2x+3|-4≤5
ANSWER #2) 3|2x+3|-4≤5
-1
0
1
3|2x+3|≤9
|2x+3|≤3
-3≤2x+3≤3
-6≤2x≤0
-3≤x≤0
-3
0
Graph Absolute Values
All absolute value functions: alx-hl+k
a- Determines if graph is narrower or wider and if reflected over x-axis
lal>1, Narrower
lal<1, Wider
a<0, Reflection over x-axis
h- Determines horizontal shifts (Left or Right)
h>0, Move right
h<0, Move left
k- Determines vertical shifts (Up or Down)
k>0, Move up
k<0, Move down
Complete the Square
Process of adding a constant to x^2+bx to make it a perfect square. (x^2+bx+c)
c= (b/2)^2
x^2+bx+c=(x+b/2)^2
Real life example: The height y (in feet) of a baseball t seconds after it is hit is given by this function:
y=-16t^2+96t+3 Find the maximum height of the baseball
1.
y=-16t^2+96t+3
2.
y=-16(t^2-6t)+3
3.
y+(-16)(?)=-16(t^2-6t+?)+3
4.
y+(-16)(9)=-16(t^2-6t+9)+3
The vertex is (3,147) so the maximum height
of the baseball is 147 feet.
Quadratic Functions
3 forms of equations
- Vertex Form
- y=a(x-h)+k where (h,k) is the vertex
- Used when the vertex and one other point is given
-
Intercept Form - y=a(x-p)(x-q) where p and q are the x-intercepts
- Used when x-intercepts and one other point is given
-
Standard form - y=ax^2+bx+c where a,b, and c are coefficients
- Used when three points are given
Quadratic Functions Examples
What is the equation of the function that …
vertex=(2,1)
x-intercepts= -1,4
goes through (0,4)
goes through (0,-4)
y=a(x-2)^2+1
y=a(x+1)(x-4)
4=a(0-2)^2+1
-4=a(1)(-4)
4=a(-2)^2+1
-4=-4a
4=4a+1
a=1
3=4a
y=(x+1)(x-4)
a=¾
y=¾(x-2)^2+1
goes through (-1,-5),(3,-1),(2,1)
-5=a(-1)^2+b(-1)+c
-5=a-b+c
c=-5-a+b
c=-5-(-1)+3
c=-1
y=-x^2+3x-1
-1=a(3)^2+b(3)+c 1=a(2)^2+b(2)+c
-1=9a+3b+c
1=4a+2b+c
-1=8a+4b-5
6=3a+3b
4=8a+4b
2=a+b
1=2a+b
2=1-a
b=1-2a
a=-1
Parabolas
axis symmetry
the focus located on the axis of symmetry
directrix
line perpendicular to the axis of symmetry
*Every point is equidistant from the directrix and the focus*
vertex
halfway between directrix and focus
p distance between focus and vertex
latus rectum 4p
COMMON MISTAKE: it is a distance so IT MUST BE POSITIVE
Parabolas
Equations when the vertex is (0,0)
x^2=4py
y^2=4px
up or down facing parabola
right or left facing parabola
if p<0 then down or left facing
parabola
if p>0 then up or right facing
parabola
focus: (0,p)
focus: (p,0)
directrix: y=-p
directrix: x=-p
axis of symmetry: x=0
axis of symmetry: y=0
Parabolas
Equations when vertex is (h,k)
y=1/4p(x-h)^2+k
x=1/4p(y-k)^2+h (a=1/4p)
directrix: y=k-p
directrix: x=h-p
focus: (h,k+p)
focus: (h+p,k)
axis of symmetry: x=h
axis of symmetry : y=k
Parabolas Examples
What is the vertex, directrix, focus, and axis of symmetry of …
x=-⅛(y^2)
vertex:(0,0) directrix: x= 2 focus: (-2,0) A of S: y=0
x-3=½(y-2)^2 vertex:(3,2) directrix: x= 5/2 focus: (7/2,3) A of S: y=2
y+1=4y-x^2 *Hint: use complete the square*
vertex: (2,3) directrix: y=13/4 focus: (2,11/4) A of S: x=2
Circles
● Equation of a circle with the center at the origin: x²+y²=R²
● Equation of a circle with center at (h,k) and radius r > 0: (x-h)²+(y-k)²=R²
EXAMPLE:
Write an equation for a circle with center (3,2) and radius 2.
*Since the center is not at the origin, we use the second equation*
(x-3)²+(y-2)²=4
EXAMPLE: Rewrite x²+y²-8x+2y+8=0 in standard form
and state the radius and center.
x²-8x+y²+2y=-8
EXAMPLE: The point (-8,5) lies on a circle whose center is (1,0). Write the
standard form of the equation of the circle.
D= √(1+8)²+(0-5)²
= √81+25
= √106
(x-1)²+(y-0)²=106
find the distance between the two points to find the radius
arrange it so that like
variables are next to each
other and move the
constant to the other side
(x²-8x+16)+(y²+2y+1)=9
(x-4)²+(y-1)²=9
Center: (4,1)
Radius: 3
use the complete
the square method
here
Ellipses
Equation for Ellipse:
c^2=a^2+b^2
Horizontal major axis: (x-h)^2/a^2+(y-k)^2/b^2=1
Vertical major axis: (x-h)^2/b^2+(y-k)^2/a^2=1
a- Vertices (Points where ellipse cut major axis)
b- Co-vertices (Points where ellipse cut minor axis)
c- Foci (Distance from center to foci)
(h,k)- Center
Ellipses Examples
Find an equation of that ellipse that has x–intercepts: ± 4,y–intercepts: ± 3, and
center at the origin.
Answer: x^2/16+y^2/9
Find the equation of the ellipse with vertices (−3,2) and (7,2) and co-vertex
(2,−1).
Answer:(x-2)^2/25+(y-2)^2/9
49(x−5)^2+25(y+2)^2=1225 and find the foci.
Answer: (5,−2±√26)
Classifying Conics
In general the graph of
Ax^2+Cy^2+Dx+Ey+F=0 is a:
When...
Circle
-4AC<0 and A=C
Parabola
-4AC=0
Ellipse
-4AC<0 and A≠C
Hyperbola
-4AC>0
Examples:
1. -10x^2+20y^2+8x-6y+7=0
2. 5x^2+2x^2+6x-2y+4=0
3. 4x^2+4y^2+2x+3y+21=0
Answers:
1. Hyperbola
2. Ellipse
3. Circle
Common Mistakes and Solutions
1. Getting Hyperbola classification confused with Ellipse
Solution: Memorization of classifications of conics
2.) The latus rectum being stated as negative
Solution: Knowing that the latus rectum is a DISTANCE so it cannot be negative
3.) Not changing the sign when dividing by a negative in an absolute value equation (< to > or vice
versa)
Solution: Remember to change the sign
4.)
Not knowing when to use “or” or “and” in inequalities
Solution: GO LA
Real Life Usage
1.)Is used in the designing of parks, buildings, monuments, etc.
2.)The path of planets around the sun
3.)Parabolic lights are used to intensify a light beam
4.)Hyperbolic and parabolic lenses and mirrors are used in telescopes
5.)Parabolic microphones intensify sound waves
Connections to Other Units
1.)Unit 3- Graphing polynomials includes graphing quadratics
2.)Unit 1&3- Factoring quadratics is used in intercept form
3.)Graphing in general- translating functions using “h” and “k” (for example
square root functions)