Adv. Intermediate Algebra – Practice problems on conics -­‐ Key 1. The course for a sailboat race includes a turnaround point marked by a stationary buoy. The sailboats follow a path that remains equidistant from the buoy and the shoreline. Suppose that the buoy is at (-­‐3, 1) and the shoreline runs along east-­‐west direction 6 miles south of the buoy. Draw a graph to show the course and the shoreline. Write an equation that describes all possible positions of your sailboat. Buoy
shoreline
Equation: y =
1
(x + 3)2 − 2 12
2. Graph the equation x = 2 (y + 1)2 – 5. Label the coordinates of the vertex, focus, endpoints of the latus rectum, and the equation of the directrix. Vertex: (-­‐5, -­‐1) Set 2 = 1/(4p) to find p. p = 1/8 or 0.125 Focus: (-­‐4.875, -­‐1) Endpoints of the latus rectum: (-­‐4.875, -­‐0.75) and (-­‐4.875, -­‐1.25) Equation of directrix: x = -­‐5.125 3. Graph the equation 3x2 + 12x – y = 0. Label the coordinates of the vertex, focus, endpoints of the latus rectum, and the equation of the directrix. Complete the square to convert 3x2 + 12x – y = 0 to standard form. 3x2 + 12x – y = 0 3x2 + 12x = y 3(x2 + 4x) = y 3(x2 + 4x + 4) = y + 12 3(x + 2) 2 -­‐ 12 = y Vertex (-­‐2, -­‐12) Parabola opens up. Set 1/(4p) = 3 to solve for p. –> p = 1/12 Focus (-­‐2, -­‐12 + 1/12); Endpts of latus rectum: (-­‐2 ± 1/6, -­‐12 + 1/12) Directrix: y = -­‐12 – 1/12 4. A cell phone tower located at the point (-­‐2, 0) sends out a signal 7 miles in every direction. a) If you are 5 miles east of the tower and start driving north, how far can you drive before you are out of range? Using Pythagorean Theorem: 52 + y2 = 72, solve for y. Y = 24 = 2 6 ≈ 4.899 b) Let (x, y) be the coordinates of a position on the coordinate plane before you are out of range. Write an equation to describe all possible positions. Equation: (x + 2)2 + y2 = 49 5. The equation x2 + y2 – 8x – 12 = 0 describes all possible positions of a cell phone user from a tower before he/she is out of range. Find the location of the tower and the maximum distance that the tower can transmit. Draw a graph to illustrate this situation. Complete the square to convert the given equation into standard form: x2 + y2 – 8x – 12 = 0 x2 – 8x + y2 = 12 x2 – 8x + 16 + y2 = 12 + 16 (x – 4)2 + y2 = 28 Center (4, 0); radius = 28 6. A "whispering room" is one with an elliptically-­‐arched ceiling. If someone stands at one focus of the ellipse and whispers something to his friend, the dispersed sound waves are reflected by the ceiling and concentrated at the other focus, allowing people across the room to clearly hear what he said. Suppose such gallery has a ceiling reaching twenty feet above the five-­‐foot-­‐high vertical walls at its tallest point (so the cross-­‐section is half an ellipse topping two vertical lines at either end), and suppose the foci of the ellipse are thirty feet apart. What is the height of the ceiling above each "whispering point"? Since the ceiling is half of an ellipse (the top half, specifically), and since the foci will be on a line
between the tops of the "straight" parts of the side walls, the foci will be five feet above the floor,
which sounds about right for people talking and listening: five feet high is close to face-high on
most adults.
Center the ellipse above the origin, so (h, k) = (0, 5). The foci are thirty feet apart, so they're 15
units to either side of the center. In particular, c = 15. Since the elliptical part of the room's crosssection is 20 ft high above the center, and since this "shorter" direction is the semi-minor axis,
then b = 20. The equation b2 = a2 – c2 gives me 400 = a2 – 225, so a2 = 625. Then the equation for the
elliptical ceiling is:
x 2 (y − 5)2
+
= 1 625
400
The height (from the ellipse's central line through its foci, up to the ceiling) will be the yvalue of the ellipse when x = 15. Substituting x = 15 into the equation and solve for y, we
get y = 21. 7. The towers of Golden Gate Bridge connecting San Francisco to Marin County are 1280 meters apart and rise 160 meters above the road. The cable between the towers has the shape of a parabola, and the cable just touches the sides of the road midway between the towers. What is the height of the cable 100 meters from a tower? Round to the nearest meter. cable
160 m
If we place the vertex of the parabola at the origin, then Equation of parabola: y = 1280 m
1 2
x or y = ax2 since we don’t need to find p for this problem. 4p
Substituting (640,160) into the equation, we can find the value of a. 160 = a(6402) –> a = 160/(6402) = 1/2560 Equation: y = 1
x 2 2560
To find the height of the cable 100 meters from a tower, substitute x = 540 into the equation. Y = 1
(540)2 ≈ 113.9m 2560
8. A fairly recent application in medicine is the use of elliptical reflectors and ultrasound to break up kidney stones. A device called a lithotripter is used to generate intense sound waves that break up the stone from outside of the body, thus avoiding surgery. To be certain that the waves do not damage other parts of the body, the reflecting property of the ellipse is used to design and correctly position the lithotripter. A lithotripter is formed by rotating the portion of an ellipse below the minor axis around the major axis. The lithotripter is 20 cm wide and 16 cm deep. If the ultrasound source is positioned at one focus of the ellipse and the kidney stone at the other, then all the sound waves will pass through the kidney stone. How far from the kidney stone should the point V on the base of the lithotripter be positioned to focus the sound waves on the kidney stone? Half of the major axis: a = 16. Half of the minor axis b = 10 Using the Pythagorean relation for an ellipse a2 = b2 + c2, distance from center to each focus is c = 156 ≈ 12.5cm Distance from V to the kidney stone is a + c = 16 + 12.5 = 28.5 cm. 9. You and your friend live 4 miles apart (on the same “east-­‐west” street) and are talking on the telephone. You hear a clap of thunder from lightning in a storm, and 18 seconds later your friend hears the thunder. Find an equation that gives the possible places where the lightning could have occurred (assume distances are measured in feet and sound travels at 1100 feet per second). From the time difference between you and your friend, 18 seconds, we can find the difference in the distances from the thunder to you and your friend. The difference between the distances is 2a in the equation, 2a = 19,800, so a = 9,900 10. Solve the system of equations algebraically. Graph the system of equations on the same coordinate plane and verify your solutions. "
$ 4 − y = x2
#
$% x 2 + y 2 + y = 8 Using substitution, we can write the second equation as (4-­‐y) + y2 + y = 8 y2 + 4 = 8 y2 = 4 y = ±2 For y = 2, substituting into the first equation, we get 4 – 2 = x2, so 2 = x2, x = ± 2 ≈ ±1.414
into the first equation, we get 4 – (-­‐2) = x2, so 6 = x2, For y = -­‐2, substituting x = ± 6 ≈ ±2.449
Solutions: ( 2,2); (− 2,2);( 6,−2); −6,−2)
For #11-­‐15, graph the conics on the same set of axes. Label each graph with the corresponding problem number. 11. Hyperbola centered at (-­‐1, 3) , one vertex at (1, 3) and one of the foci at (-­‐ 1+ 13 , 3) for 0 ≤ y ≤ 3. 12. –8(y – 6) = (x + 1)2 for 4 ≤ y ≤ 6 only. 13. 9x2 + 4y2 + 18x – 24y + 9 = 0 14. (x + 1)2 + (y – 7) 2 = 1 €
15. Write an equation (in standard form) of a conic and add its graph to the picture.