PRE-CALCULUS MIDTERM REVIEWER CONICS Circle is the set of points in a plane that are equidistant from a fixed point called center. The distance from the center to any point on the circle is called the radius. Standard Equation of the circle: πΆπππ‘ππ ππ‘ (0,0): π₯ 2 + π¦ 2 = π 2 πΆπππ‘ππ ππ‘ (β, π): (π₯ − β)2 + (π¦ − π)2 = π 2 Parabola is the set of points in a plane equidistant from a fixed point called focus and a fixed line called directrix. Standard Equation of the parabola: ππππ‘ππ₯ ππ‘ (0,0): π₯ 2 = 4ππ¦ ππ π¦ 2 = 4ππ₯ ππππ‘ππ₯ ππ‘ (β, π): (π₯ − β)2 = 4π(π¦ − π) ππ (π¦ − π)2 = 4π(π₯ − β) Ellipse is the set of points in a plane whose sum of distances from two fixed points called foci is a constant. Standard Equation of the ellipse: 2 π₯ π¦2 π₯2 π¦2 πΆπππ‘ππ ππ‘ (0,0): 2 + 2 = 1 ππ 2 + 2 = 1, π€βπππ π > π πππ π, π > 0 π π π π πΆπππ‘ππ ππ‘ (β, π): (π₯ − β)2 (π¦ − π)2 (π₯ − β)2 (π¦ − π)2 + = 1 ππ + = 1, π€βπππ π > π πππ π, π > 0 π2 π2 π2 π2 Hyperbola is the set of points in a plane whose absolute value of the difference of distances from two fixed points is a constant. Standard Equation of the hyperbola: π₯2 π¦2 π¦2 π₯2 πΆπππ‘ππ ππ‘ (0,0): 2 − 2 = 1 ππ 2 − 2 = 1, π€βπππ π, π > 0 π π π π πΆπππ‘ππ ππ‘ (0,0): (π₯ − β)2 (π¦ − π)2 (π¦ − π)2 (π₯ − β)2 − = 1 ππ − = 1, π2 π2 π2 π2 π€βπππ π, π > 0 • In solving for the value of r of a circle, if r>0, then there would be a legitimate value of r of a circle. If r=0, then there would be a point circle. If r<0, then there would be an empty set, no circle is formed. • The point between the focus and the directrix is called the vertex. It is the point on the parabola whose distance from the focus is the shortest. • A parabola with an equation π₯ 2 = 4ππ¦, the line segment that is perpendicular to the π¦ − ππ₯ππ and passes through the focus is called latus rectum, also known as focal width. • In a parabola, if π > π the parabola opens upward or to the right. If π < π the parabola opens downward or to the left. • In an ellipse, the axis that contains the center, foci, and vertices is the major axis. The axis that is perpendicular to the major axis is the minor axis. The major axis is a line segment that connects the vertices while the minor axis is a line segment that connects the covertices. • In solving problems on ellipse, the value of a is oftentimes greater than b. if a=b, then a circle is formed. • • If π πππ π are the lengths of the semi-major and semi-minor axes, respectively, then the distance of the center to the focus is denoted by c and may be obtained from the equation ππ = ππ + ππ . The following properties are necessary in graphing an ellipse: a. Coordinates of the center b. Values of a and b. c. Orientation of the principal axis (horizontal or vertical). • A hyperbola has two axes – the transverse axis and the conjugate axis, whose lengths are ππ and ππ, respectively. The transverse axis connects the vertices while the conjugate axis connects the covertices. • If π πππ π are the lengths of the semi-transverse and semi-conjugate axes, respectively, then the distance of the center to the focus is denoted by c and may be obtained from the equation ππ = ππ + ππ . Sample Problems with Solutions: 1. Transform 3π₯ 2 − 12π₯ + 4π¦ 2 − 8π¦ + 16 = 0 into a standard equation and find its properties. Given the equation above, we cannot find the properties of the ellipse since the general equation results to an equation of a degenerate conic. 2. The reflector of a flashlight is in the shape of a parabolic surface. The casting has a diameter of 4 inches and has depth of 2 inches. If the lightbulb represents the focus of the parabolic reflector, how far from the vertex should the lightbulb be placed? What is the general equation that represents the parabola? Since the vertex of the parabola is at the origin and p=1/2, we can say that the lightbulb should be placed 0.5 inches from the vertex. The general equation of the parabola is 3. Find the general equation of the ellipse with center located at (-3,1) and has lengths of the vertical major and minor axes 10 units and 8 units, respectively. From the given problem, since the major axis is vertical, we can say that the standard equation of the ellipse is (π₯−β)2 π2 + (π¦−π)2 π2 = 1. Now, given that the center is located at (-3,1), with major axis that has a length of 10 units and minor axis that has a length of 8 units, we have (π₯ + 3)2 (π¦ − 1)2 + =1 16 25 From the standard equation, we can determine the general equation as follows (π₯ + 3)2 (π¦ − 1)2 + =1 16 25 Answered Exercises: 1. Find an equation of a circle whose center is at the origin and whose radius is 4. π¨πππππ: ππ + ππ = ππ 2. Find and equation of a circle whose center is at (2, -5) and whose radius is 3. π¨πππππ: (π − π)π + (π + π)π = π 3. Determine the center and radius of the circle having an equation π₯ 2 + π¦ 2 − 10π₯ + 6π¦ − 2 = 0. π¨πππππ: πͺ(π, −π); π = π 4. Find the general form of the equation of the circle whose diameter has endpoints at (5,-6) and (-3,-6). π¨πππππ: πͺ(π, −π); π = π; ππ + ππ − ππ + πππ + ππ = π 5. The equation of a parabola is π¦ 2 = 8π₯. Find the coordinates of the vertex, focus, endpoints of the latus rectum, and the equation of the directrix. π¨πππππ: π½(π, π); π(π, π); π³π(π, π)&π³π(π, −π); πππ π = −π 6. The equation of a parabola is (π₯ + 2)2 = −12(π¦ − 5). Find the coordinates of the vertex, focus, endpoints of the latus rectum, and the equation of the directrix. π¨πππππ: π½(−π, π); π(−π, π); π³π(−π, π)&π³π(π, π); πππ π = π 7. Find an equation of a parabola with π(0,0)πππ πΉ(0, −4). π¨πππππ: ππ = −πππ 8. Find the general equation of a parabola with πΉ(3,1)πππ ππππππ‘πππ₯ ππ π¦ = −2. π¨πππππ: ππ − ππ − ππ + π = π 9. A particular cell phone tower is designed to service a 15-mile radius. The tower is located at (-4,4) on a coordinate plane whose units represent miles. What is the general equation of the outer boundary of the region serviced by the tower? π¨πππππ: ππ + ππ + ππ − ππ − πππ = π 10. A cable hangs in a parabolic arc between two poles 100 feet apart. The poles are 30 feet high and the lowest point on the suspended cable is 5 feet above the ground. Find the equation of the arc if the vertex is the lowest point of the cable. Also, find the height of the cable at a point 10 feet from one of the poles. π¨πππππ: ππ = πππ(π − π); ππ ππππ 11. The equation of the ellipse is (π₯−1)2 16 + (π¦+3)2 9 = 1. Find the coordinates of the center and the foci. π¨πππππ: πͺ(π, −π); π(π ± √π, −π) πππππ π = π, π = π, πππ π = √π 12. Find an equation of the ellipse having πΉ1(0,4) πππ πΉ2(0, −4), and one endpoint of the minor axis is at (-3,0) ππ ππ π¨πππππ: + = π ππ ππππ + πππ = πππ π ππ 13. The orbit of Earth around the sun is in the shape of an ellipse, where the Sun is at one of the foci, and whose length of the major axis is 185.8 million miles. If the distance of the sun from the center of the orbit is 1.58 million miles, find the least distance and the greatest distance of Earth from the Sun. π¨πππππ: πππππ π πππππππ − ππ. ππ πππππππ πππππ; ππππππππ π πππππππ − ππ. ππ πππππππ πππππ 14. Given the equation of the hyperbola 16π₯ 2 − 9π¦ 2 = 144, find the coordinates of the center, vertices, endpoints of the conjugate axis, foci, and length of transverse and conjugate axes. π¨πππππ: πͺ(π, π); π½π(−π, π)π½π(π, π); π΄π(π, π)π΄π(π, −π); ππ(−π, π)ππ(π, π); ππ = π; ππ = π (π¦−2)2 (π₯+6)2 15. Given the equation of the hyperbola − = 1, find the coordinates of the center, vertices, 45 36 endpoints of the conjugate axis, foci, and length of transverse and conjugate axes. π¨πππππ: πͺ(−π, π); π½π(−π, π + π√π)π½π(−π, π − π√π); π΄π(−ππ, π)π΄π(π, π); ππ(−π, ππ)ππ(−π, −π); ππ = π√π; ππ = ππ 16. At an air show, a stunt plane dives along a hyperbolic path whose vertex is directly over a stage with a height of 22 meters. Exactly 15 meters above the stage is the center of the hyperbolic path. If the plane’s flight path can be modeled by the hyperbola 64π¦ 2 − 3600π₯ 2 = 230,400, what is the minimum distance of the plane from the ground? From the rooftop of the building? Assume x and y are in meters π¨πππππ: ππ ππππππ & ππ ππππππ
