# Unit 2 Conic Sections By Areeb Rahim Vibha

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. (0, 3), (x, 5); d= 2√ 10= √(x-0)²+(5 -3)² (2√ 10)²= (√x²+4)² 40= x²+4 36= x² ± 6= x AC= √(6 --2)²+(3 --5)² = √ 64+64 = √ 128 = 8√ 2 EXAMPLE: Find the midpoint of the segment joining (2, 7), (8, 2) 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 EXAMPLE: |x-9|+5=10 *when the absolute value equals a negative number, it is no solution* |x-9|=5 x-9=5 or x-9=-5 x=14 or x=4 isolate the absolute value *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|6 x-1|+5<3 -2|6 x-1|<-2 |6 x-1|>1 6 x-1>1 or 6 x-1<1 6 x>2 or 6 x<0 x>⅓ or x<0 PRACTICE #1) Solve the inequality, don’t graph: -7+2|3 x-2|≥ 9 ANSWER #1) -7+2|3 x-3|≥ 9 2|3 x-2|≥ 16 |3 x-2|≥ 8 3 x-2≥ 8 or 3 x-2≤-8 3 x≥ 10 or 3 x≤-6 x≥ 10/3 or x≤-2 PRACTICE #2) Solve and graph the inequality: 3|2 x+3|-4≤ 5 ANSWER #2) 3|2 x+3|-4≤ 5 -1 0 1 3|2 x+3|≤ 9 |2 x+3|≤ 3 -3≤ 2 x+3≤ 3 -6≤ 2 x≤ 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=-16 t^2+96 t+3 Find the maximum height of the baseball 1. y=-16 t^2+96 t+3 2. y=-16(t^2 -6 t)+3 3. y+(-16)(? )=-16(t^2 -6 t+? )+3 4. y+(-16)(9)=-16(t^2 -6 t+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=-4 a 4=4 a+1 a=1 3=4 a 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+3 x-1 -1=a(3)^2+b(3)+c 1=a(2)^2+b(2)+c -1=9 a+3 b+c 1=4 a+2 b+c -1=8 a+4 b-5 6=3 a+3 b 4=8 a+4 b 2=a+b 1=2 a+b 2=1 -a b=1 -2 a 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 4 p COMMON MISTAKE: it is a distance so IT MUST BE POSITIVE

Parabolas Equations when the vertex is (0, 0) x^2=4 py y^2=4 px 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=1/4 p(x-h)^2+k x=1/4 p(y-k)^2+h (a=1/4 p) 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=4 y-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+y²-8 x+2 y+8=0 in standard form and state the radius and center. x²-8 x+y²+2 y=-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 (x²-8 x+16)+(y²+2 y+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 -4 AC<0 and A=C Parabola -4 AC=0 Ellipse -4 AC<0 and A≠C Hyperbola -4 AC>0 Examples: 1. -10 x^2+20 y^2+8 x-6 y+7=0 2. 5 x^2+2 x^2+6 x-2 y+4=0 3. 4 x^2+4 y^2+2 x+3 y+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)

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