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๐ŸฌHonors Algebra II Unit 10 Review

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10.1 Parabolas and Circles

๐ŸฌHonors Algebra II
Unit 10 Review

10.1 Parabolas and Circles

Written by the Fiveable Content Team โ€ข Last updated September 2025
Written by the Fiveable Content Team โ€ข Last updated September 2025
๐ŸฌHonors Algebra II
Unit & Topic Study Guides

Parabolas and circles are key players in the conic sections family. They're everywhere, from satellite dishes to car wheels. Understanding their shapes, equations, and real-world uses is crucial for mastering this topic.

These curves have unique features that make them special. Parabolas have a vertex and symmetry, while circles are all about the center and radius. Graphing and finding equations for both shapes involves similar steps, making them a perfect pair to study together.

Parabolas and circles: Components and characteristics

Parabola fundamentals

  • A parabola is a U-shaped curve that is symmetrical and has a single turning point called the vertex
  • Parabolas extend infinitely in one direction
  • The vertex form of a parabola is y=a(xโˆ’h)2+ky=a(x-h)^2+k, where (h,k)(h,k) is the vertex and aa determines the direction and width of the parabola
  • Examples of parabolic shapes include satellite dishes and the Gateway Arch in St. Louis

Parabola symmetry and key features

  • Parabolas have an axis of symmetry, a vertical line that passes through the vertex and divides the parabola into two mirror images
  • The equation of the axis of symmetry is x=hx=h
  • The directrix is a horizontal line perpendicular to the axis of symmetry
  • The focus is a point on the axis of symmetry that is equidistant from the vertex as the directrix

Circle fundamentals

  • A circle is a round plane figure whose boundary consists of points equidistant from the center
  • The standard form of a circle is (xโˆ’h)2+(yโˆ’k)2=r2(x-h)^2+(y-k)^2=r^2, where (h,k)(h,k) is the center and rr is the radius
  • Examples of circular objects include wheels, coins, and pizzas

Circle symmetry and key features

  • Circles are symmetrical about any diameter, a line segment that passes through the center and has its endpoints on the circle
  • The chord is a line segment that connects any two points on the circle's circumference
  • The central angle is an angle formed by two radii drawn to the endpoints of an arc
  • The inscribed angle is an angle formed by two chords that share an endpoint on the circle's circumference

Graphing parabolas and circles

Graphing parabolas

  • To graph a parabola, plot the vertex (h,k)(h,k), then plot additional points on either side of the vertex using the equation
  • Connect the plotted points to form the U-shape
  • The sign of aa in the vertex form determines if the parabola opens up (a>0)(a>0) or down (a<0)(a<0)
  • The absolute value of aa affects the width of the parabola (larger โˆฃaโˆฃ|a| means narrower parabola)

Parabola graphing techniques

  • Graphing parabolas requires a t-chart to solve for ordered pairs, with x-values symmetrical around the axis of symmetry
  • Choose x-values that are equidistant from hh, such as hโˆ’1h-1, hh, and h+1h+1, to ensure symmetry
  • Substitute the x-values into the equation to find the corresponding y-values
  • Plot the ordered pairs and connect them to form the parabola

Graphing circles

  • To graph a circle, plot the center (h,k)(h,k), then plot four points rr units up, down, left, and right from the center
  • Connect the four plotted points to form the circle
  • The general form of a circle is x2+y2+Ax+By+C=0x^2+y^2+Ax+By+C=0
  • To graph a circle in general form, convert it to standard form by completing the square for both xx and yy

Circle graphing techniques

  • Identify the coefficients AA and BB of the xx and yy terms, respectively
  • Divide AA and BB by 2 and square the results to find hh and kk
  • Substitute hh and kk into the general form and simplify to find CC
  • Use the Pythagorean theorem to solve for the radius: r=h2+k2โˆ’Cr=\sqrt{h^2+k^2-C}
  • Plot the center and points, then connect to form the circle

Equations of parabolas and circles

Determining parabola equations

  • To find the equation of a parabola, identify the vertex (h,k)(h,k) and a point (x,y)(x,y) on the curve
  • Substitute the coordinates into y=a(xโˆ’h)2+ky=a(x-h)^2+k and solve for aa
  • The axis of symmetry and directrix can also be used to determine the equation of a parabola
  • Given the focus (h,k+p)(h,k+p) and directrix y=kโˆ’py=k-p, the equation is y=14p(xโˆ’h)2+ky=\frac{1}{4p}(x-h)^2+k

Determining circle equations

  • To find the equation of a circle, identify the center (h,k)(h,k) and radius rr
  • Substitute the values into (xโˆ’h)2+(yโˆ’k)2=r2(x-h)^2+(y-k)^2=r^2
  • Given the general form x2+y2+Ax+By+C=0x^2+y^2+Ax+By+C=0, convert to standard form to identify the center and radius
  • The equation of a circle can also be determined using the midpoint and distance formulas if given the endpoints of a diameter

Converting between forms

  • To convert from standard form to general form, expand the squared binomials and simplify
  • To convert from general form to standard form, complete the square for both xx and yy
  • Vertex form is specific to parabolas and is not used for circles
  • Converting between forms is often necessary to extract relevant information for problem-solving

Real-world applications of parabolas and circles

Parabola applications

  • Many real-world situations can be modeled by parabolas, such as trajectories of objects (footballs, basketballs), satellite dishes, suspension bridges, and certain curves in nature
  • The vertex often represents the maximum or minimum value of the parabola in a particular context
  • Example: The height hh of a ball thrown upward with an initial velocity of 30 ft/s from a height of 5 ft can be modeled by h=โˆ’16t2+30t+5h=-16t^2+30t+5, where tt is time in seconds
  • The vertex (0.9375,19.0625)(0.9375, 19.0625) represents the maximum height of the ball at 0.9375 seconds after being thrown

Circle applications

  • Circles can be used to model the wheels of a car, traffic roundabouts, center-pivot irrigation systems, and the distance from a central location
  • Example: A center-pivot irrigation system rotates around a central point, watering a circular area. If the system is 1,320 feet long, find the area it irrigates.
  • The area can be found using A=ฯ€r2A=\pi r^2, where r=1,320r=1,320. The system irrigates approximately 5,471,136 square feet.

Problem-solving strategies

  • Real-world problems often require setting up an equation based on given information and constraints, then solving that equation for a specific value
  • Identify the key components of the problem, such as the vertex, center, radius, or points on the curve
  • Determine which form of the equation (vertex, standard, general) is most appropriate for the given information
  • Translate the problem into an equation by substituting the known values
  • Solve the equation for the desired quantity, such as a coordinate, dimension, or optimum value
  • Parabolas and circles may need to be translated between different forms to extract the relevant information to answer the question