7.3 Parabolas. 7.3 Parabolas 505

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1 7. Parabolas 0 7. Parabolas We have alread learned that the graph of a quadratic function f() = a + b + c (a 0) is called a parabola. To our surprise and delight, we ma also define parabolas in terms of distance. Definition 7.. Let F be a point in the plane and D be a line not containing F. A parabola is the set of all points equidistant from F and D. The point F is called the focus of the parabola and the line D is called the directri of the parabola. Schematicall, we have the following. F V Each dashed line from the point F to a point on the curve has the same length as the dashed line from the point on the curve to the line D. The point suggestivel labeled V is, as ou should epect, the verte. The verte is the point on the parabola closest to the focus. We want to use onl the distance definition of parabola to derive the equation of a parabola and, if all is right with the universe, we should get an epression much like those studied in Section.. Let p denote the directed distance from the verte to the focus, which b definition is the same as the distance from the verte to the directri. For simplicit, assume that the verte is (0, 0) and that the parabola opens upwards. Hence, the focus is (0, p) and the directri is the line = p. Our picture becomes D (, ) (0, p) (0, 0) = p (, p) From the definition of parabola, we know the distance from (0, p) to (, ) is the same as the distance from (, p) to (, ). Using the Distance Formula, Equation., we get We ll talk more about what directed means later.

2 06 Hooked on Conics ( 0) + ( p) = ( ) + ( ( p)) + ( p) = ( + p) + ( p) = ( + p) square both sides + p + p = + p + p epand quantities = p gather like terms Solving for ields = p, which is a quadratic function of the form found in Equation. with a = p and verte (0, 0). We know from previous eperience that if the coefficient of is negative, the parabola opens downwards. In the equation = p this happens when p < 0. In our formulation, we sa that p is a directed distance from the verte to the focus: if p > 0, the focus is above the verte; if p < 0, the focus is below the verte. The focal length of a parabola is p. If we choose to place the verte at an arbitrar point (h, k), we arrive at the following formula using either transformations from Section.7 or re-deriving the formula from Definition 7.. Equation 7.. The Standard Equation of a Vertical a Parabola: The equation of a (vertical) parabola with verte (h, k) and focal length p is ( h) = p( k) If p > 0, the parabola opens upwards; if p < 0, it opens downwards. a That is, a parabola which opens either upwards or downwards. Notice that in the standard equation of the parabola above, onl one of the variables,, is squared. This is a quick wa to distinguish an equation of a parabola from that of a circle because in the equation of a circle, both variables are squared. Eample 7... Graph ( + ) = 8( ). Find the verte, focus, and directri. Solution. We recognize this as the form given in Equation 7.. Here, h is + so h =, and k is so k =. Hence, the verte is (, ). We also see that p = 8 so p =. Since p < 0, the focus will be below the verte and the parabola will open downwards. 6

3 7. Parabolas 07 The distance from the verte to the focus is p =, which means the focus is units below the verte. From (, ), we move down units and find the focus at (, ). The directri, then, is units above the verte, so it is the line =. Of all of the information requested in the previous eample, onl the verte is part of the graph of the parabola. So in order to get a sense of the actual shape of the graph, we need some more information. While we could plot a few points randoml, a more useful measure of how wide a parabola opens is the length of the parabola s latus rectum. The latus rectum of a parabola is the line segment parallel to the directri which contains the focus. The endpoints of the latus rectum are, then, two points on opposite sides of the parabola. Graphicall, we have the following. the latus rectum F V It turns out that the length of the latus rectum, called the focal diameter of the parabola is p, which, in light of Equation 7., is eas to find. In our last eample, for instance, when graphing ( + ) = 8( ), we can use the fact that the focal diameter is 8 = 8, which means the parabola is 8 units wide at the focus, to help generate a more accurate graph b plotting points units to the left and right of the focus. Eample 7... Find the standard form of the parabola with focus (, ) and directri =. Solution. Sketching the data ields, D The verte lies on this vertical line midwa between the focus and the directri No, I m not making this up. Consider this an eercise to show what follows.

4 08 Hooked on Conics From the diagram, we see the parabola opens upwards. (Take a moment to think about it if ou don t see that immediatel.) Hence, the verte lies below the focus and has an -coordinate of. To find the -coordinate, we note that the distance from the focus to the directri is ( ) =, which means the verte lies units (halfwa) below the focus. Starting at (, ) and moving down / units leaves us at (, /), which is our verte. Since the parabola opens upwards, we know p is positive. Thus p = /. Plugging all of this data into Equation 7. give us ( ) ( ( ) = ( ( ) = 0 + ) ( )) If we interchange the roles of and, we can produce horizontal parabolas: parabolas which open to the left or to the right. The directrices of such animals would be vertical lines and the focus would either lie to the left or to the right of the verte, as seen below. D V F Equation 7.. The Standard Equation of a Horizontal Parabola: The equation of a (horizontal) parabola with verte (h, k) and focal length p is ( k) = p( h) If p > 0, the parabola opens to the right; if p < 0, it opens to the left. plural of directri

5 7. Parabolas 09 Eample 7... Graph ( ) = ( + ). Find the verte, focus, and directri. Solution. We recognize this as the form given in Equation 7.. Here, h is + so h =, and k is so k =. Hence, the verte is (, ). We also see that p = so p =. Since p > 0, the focus will be the right of the verte and the parabola will open to the right. The distance from the verte to the focus is p =, which means the focus is units to the right. If we start at (, ) and move right units, we arrive at the focus (, ). The directri, then, is units to the left of the verte and if we move left units from (, ), we d be on the vertical line =. Since the focal diameter is p =, the parabola is units wide at the focus, and thus there are points 6 units above and below the focus on the parabola As with circles, not all parabolas will come to us in the forms in Equations 7. or 7.. If we encounter an equation with two variables in which eactl one variable is squared, we can attempt to put the equation into a standard form using the following steps. To Write the Equation of a Parabola in Standard Form. Group the variable which is squared on one side of the equation and position the nonsquared variable and the constant on the other side.. Complete the square if necessar and divide b the coefficient of the perfect square.. Factor out the coefficient of the non-squared variable from it and the constant. Eample 7... Consider the equation =. Put this equation into standard form and graph the parabola. Find the verte, focus, and directri. Solution. We need a perfect square (in this case, using ) on the left-hand side of the equation and factor out the coefficient of the non-squared variable (in this case, the ) on the other.

6 0 Hooked on Conics = + = = complete the square in onl ( + ) = factor ( + ) = 8( ) Now that the equation is in the form given in Equation 7., we see that h is so h =, and k is + so k =. Hence, the verte is (, ). We also see that p = 8 so that p =. Since p < 0, the focus will be the left of the verte and the parabola will open to the left. The distance from the verte to the focus is p =, which means the focus is units to the left of, so if we start at (, ) and move left units, we arrive at the focus (, ). The directri, then, is units to the right of the verte, so if we move right units from (, ), we d be on the vertical line =. Since the focal diameter is p is 8, the parabola is 8 units wide at the focus, so there are points units above and below the focus on the parabola. 6 In studing quadratic functions, we have seen parabolas used to model phsical phenomena such as the trajectories of projectiles. Other applications of the parabola concern its reflective propert which necessitates knowing about the focus of a parabola. For eample, man satellite dishes are formed in the shape of a paraboloid of revolution as depicted below.

7 7. Parabolas Ever cross section through the verte of the paraboloid is a parabola with the same focus. To see wh this is important, imagine the dashed lines below as electromagnetic waves heading towards a parabolic dish. It turns out that the waves reflect off the parabola and concentrate at the focus which then becomes the optimal place for the receiver. If, on the other hand, we imagine the dashed lines as emanating from the focus, we see that the waves are reflected off the parabola in a coherent fashion as in the case in a flashlight. Here, the bulb is placed at the focus and the light ras are reflected off a parabolic mirror to give directional light. F Eample 7... A satellite dish is to be constructed in the shape of a paraboloid of revolution. If the receiver placed at the focus is located ft above the verte of the dish, and the dish is to be feet wide, how deep will the dish be? Solution. One wa to approach this problem is to determine the equation of the parabola suggested to us b this data. For simplicit, we ll assume the verte is (0, 0) and the parabola opens upwards. Our standard form for such a parabola is = p. Since the focus is units above the verte, we know p =, so we have = 8. Visuall, units wide (6, )? 6 6 Since the parabola is feet wide, we know the edge is 6 feet from the verte. To find the depth, we are looking for the value when = 6. Substituting = 6 into the equation of the parabola ields 6 = 8 or = 6 8 = 9 =.. Hence, the dish will be. feet deep.

8 Hooked on Conics 7.. Eercises In Eercises - 8, sketch the graph of the given parabola. Find the verte, focus and directri. Include the endpoints of the latus rectum in our sketch.. ( ) = 6. ( + ) 7 ( ) = +. ( ) = ( + ). ( + ) =. ( ) = ( + ) 6. ( + ) = 0( ) 7. ( ) = 8( ) 8. ( + ) ( ) = In Eercises 9 -, put the equation into standard form and identif the verte, focus and directri = = = = = = 0 In Eercises - 8, find an equation for the parabola which fits the given criteria.. Verte (7, 0), focus (0, 0) 6. Focus (0, ), directri = 7. Verte ( 8, 9); (0, 0) and ( 6, 0) are points on the curve 8. The endpoints of latus rectum are (, 7) and (, 7) 9. The mirror in Carl s flashlight is a paraboloid of revolution. If the mirror is centimeters in diameter and. centimeters deep, where should the light bulb be placed so it is at the focus of the mirror? 0. A parabolic Wi-Fi antenna is constructed b taking a flat sheet of metal and bending it into a parabolic shape. If the cross section of the antenna is a parabola which is centimeters wide and centimeters deep, where should the receiver be placed to maimize reception?. A parabolic arch is constructed which is 6 feet wide at the base and 9 feet tall in the middle. Find the height of the arch eactl foot in from the base of the arch.. A popular novelt item is the mirage bowl. Follow this link to see another startling application of the reflective propert of the parabola.. With the help of our classmates, research spinning liquid mirrors. To get ou started, check out this website. This shape is called a parabolic clinder.

9 7. Parabolas 7.. Answers. ( ) = 6 Verte (, 0) Focus (, ) Directri = Endpoints of latus rectum (, ), (, ) ( + ) 7 ( ) = + Verte ( 7, ) Focus ( 7, ) Directri = Endpoints of latus rectum ( 0, ), (, ). ( ) = ( + ) Verte (, ) Focus ( 6, ) Directri = 0 Endpoints of latus rectum ( 6, 8), ( 6, )

10 Hooked on Conics. ( + ) = Verte (0, ) Focus (, ) Directri = Endpoints of latus rectum (, ), (, 6) ( ) = ( + ) Verte (, ) Focus (, ) Directri = Endpoints of latus rectum (, ), (, ) 6. ( + ) = 0( ) Verte (, ) Focus (, 0) Directri = 0 Endpoints of latus rectum (, 0), (8, 0) ( ) = 8( ) Verte (, ) Focus (, ) Directri = Endpoints of latus rectum (, ), (, )

11 7. Parabolas 8. ( + ) ( ) = Verte ( 9, ) Focus (, ) Directri = Endpoints of latus rectum (, ), (, ) 9. ( ) = 7( ) Verte (, ) Focus (, ) Directri =. ( + ) = 8( 6) Verte (, 6) Focus (, 8) Directri = 0. ( + ) = ( ) Verte (, ) Focus (, 9 ) 0 Directri = 0. ( + ) = ( 0) Verte (0, ) Focus ( 79 8, ) Directri = 8 8. ( ) = ( ). ( 9 ) = ( ) Verte (, ) Verte (, 9 ) Focus (, ) Focus (, 9 ) Directri = Directri = 7. = 8( 7) 6. ( ) = 0 ( ) 7. ( + 8) = 6 9 ( + 9) 8. ( ) = 6 ( + 7 ) or ( ) = 6 ( + ) 9. The bulb should be placed 0.6 centimeters above the verte of the mirror. (As verified b Carl himself!) 0. The receiver should be placed.06 centimeters from the verte of the cross section of the antenna.. The arch can be modeled b = ( 9) or = 9. One foot in from the base of the arch corresponds to either = ±, so the height is = 9 (±) = feet.