INSTRUCTORS Raffi Hovasapian John Zhu

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• ## Related Books

 0 answersPost by James Xie on June 21, 2012For the x^2 4x 1 question, couldn't the answer be x=rad(3)-2 ? (or x=rad(3) 2)?

### Conic Sections

• Parabolas: and
• Circles:
• Ellipses:
• Hyperbolas: and
• Completing the square
• To create an expression ready to “unfoil”
• Transform to standard conic sections form

### Conic Sections

Graph f(x) = x2
• f(0) = 0
• f(1) = 1 , f(−1) = 1
• f(2) = 4 , f(−2) = 4
• f(3) = 9 , f(−3) = 9
Graph y = (x−2)2 + 4
• This is the same as the graph of y = x2, but shifted up 4 units, and 2 to the right
Graph y = x2 + 2x + 3
• y = x2 + 2x + 3
• = x2 + 2x + 1 + 2
• = (x + 1)2 + 2
• y = x2 shifted up 2, and 1 to the left.
Graph x2 + y2 = cos2 x + sin2 x
• x2 + y2 = cos2 x + sin2 x
• x2 + y2 = 1
• y2 = 1 − x2
• y = ±√{1 − x2}
• The domain must be such that we're taking the sqrt of a non-negative number. So 1 − x2≥ 0.
• 1 − x2≥ 0
• x2≤ 1
• −1 ≤ x ≤ 1
• We can take advantage of some symmetry here. Because of the plus or minus, we know that every x will have a positive and negative solution with the same absolute value. In other words, f(x) = c and -c. Also, this is an even function, so we have symmetry about the y-axis. We only need to find values for one quadrant, and just mirror those values.
• f(0) = 1
• f(.5) ≈ 0.9
• f(.75) ≈ .7
• f(1) = 0
Graph x2 − y2 = 1
• This is the equation of a hyperbola. As |x| gets large, the equation approaches the following two asymptotes
• x2 − y2 = 0
• y2 = x2
• y = ±x
• Determine the y-intercepts
• 0 − y2 = 1
y = √{−1}
• That's an imaginary number. This means that at no point does the graph cross x = 0. Finding the x-intercepts...
• x2 − 0 = 1
x = ±1
Graph (x − 2)2 + (y − 1)2 = 4
• That's the equation of a circle with its center moved 2 to the right and up one unit with radius √4 = 2.
Graph (x − 2)2 − (y − 1)2 = 4
• Asymptote lines are
(x − 2)2 − (y − 1)2 = 0
(y − 1)2 = (x − 2)2
y − 1 = ±(x − 2)
y = 1 ±(x − 2)
y = x − 1, 3 − x
• y-intercepts
(0 − 2)2 − (y − 1)2 = 4
4 − (y − 1)2 = 4
(y − 1)2 = 0
y = 1
• x-intercepts (x − 2)2 − (0 − 1)2 = 4
(x − 2)2 − 1 = 4
(x − 2)2 = 3
x − 2 = ±√3
x = 2 ±√3
x = 2 + √3, 2 − √3
Graph ([x/2])2 + ([y/3])2 = 1
• This is the equation of an ellipse extending out 2 on the x-axis, and 3 on the y-axis from the center. In this case, the center is at 0,0.
Graph ([(x + 1)/3])2 + ([(y − 5)/2])2 = 16
• This is the equation of an ellipse with its center at -1, 5. To find out how far out the ellipse goes on each respective sides, plug in the center values.
• ([(−1 + 1)/3])2 + ([(y − 5)/2])2 = 16
([(y − 5)/2])2 = 16
[(y − 5)/2] = ±4
y − 5 = ±8
y = 5 ±8
y = 13, −3
• ([(x + 1)/3])2 + ([(5 − 5)/2])2 = 16
([(x + 1)/3])2 = 16
[(x + 1)/3] = ±4
x + 1 = ±12
x = −1 ±12
x = −13, 11
Graph x2 + 6x = −y2 + 4y
• Completing the squares:
x2 + 6x = −y2 + 4y
x2 + 6x + y2 − 4y = 0
x2 + 6x + 9 − 9 + y2 − 4y + 4 − 4 = 0
(x + 3)2 − 9 + (y − 2)2 − 4 = 0
(x + 3)2 + (y − 2)2 = 13
• This is a circle with center [-3, 2] and radius √{13}

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Conic Sections

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Types of Conic Sections 0:06
• Parabolas
• Circles
• Ellipses
• Hyperbolas
• Complete the Square 6:40
• Example: Conic Sections 9:08
• Example 2: Conic Sections 10:59
• Example 3: Graph Conic Sections 12:21