HYPERBOLAS BY: Valencia Williams
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1 HYPERBOLAS BY: Valencia Williams A hyperbola is the set of all points (x, y) in a plane such that the absolute value of the difference of the distance from 2 fixed points (foci) is a positive constant. The equation of a hyperbola is: The equation for vertical transverse axis: (y k)2 The equation for horizontal transverse axis: (x h)2 a 2 a 2 (x h)2 b 2 = 1 (y k)2 b 2 = 1 The components of a hyperbola are: Vertical Transverse Axis The center: (h, k) The vertices :( h, k ±a) The foci (focus) :( h, k±c) The asymptote equation: y = k ± a (x h) b The equation relating a, b, and c: c 2 = a 2 + b 2 To remember the equation for vertical transverse axis always remember that y comes first in the equation. Horizontal Transverse Axis The center: (h, k)
2 The vertices: (h±a, k) The foci (focus): (h±c, k) The asymptote equation: y = k ± b (x h) a The equation relating a, b, and c: c 2 = a 2 + b 2 To remember the equation for the horizontal transverse axis always remember that x comes first in the equation. The red dot in the center is represented by the center. The purple dots represent the vertex and you can determine them by the given a value. You start at the center and count out for horizontal and count upward/downward for vertical. The green dots represent the foci/focus which is found by adding or subtracting your c value from your k for vertical and you h for horizontal. The blue lines represent the asymptote of the hyperbola. You draw it by the drawing an outward half circle through the vertex for horizontal and upward/downward half circle through the vertex for vertical. You can see The Algebra and Trigonometry Book pages for more information on Hyperbolas. Example Problem #1 (x + 3) 2 (y + 1)2 = Find all the components of the hyperbola equation. The first step is to determine if the equation is vertical transverse or horizontal transverse. o It is horizontal transverse axis because x comes first.
3 The next step is to determine the center which is in the equation. o Center (-3, -1) It is negative because in the original equation it is a subtract sign so if there is an addition sign that means the number is negative. Now that you know you have a horizontal transverse and the center has been determined. The a and b values have to be found. In the original equation the a and b values are squared. So unsquared the a and b value above in the given equation. o 4 = 2 and 9 = 3 A value= 2 B value= 3 Next you need to find the vertices of the given equation. The formula to find the vertices is (h±a, k). o (-3+2, -1) = (-1, -1) o (-3-2, -1)= (-5, -1) The c value has to be determined in order to find the foci/focus. The formula to determine the c value isc 2 = a 2 + b 2. o c 2 = c 2 = c 2 = 13 c 2 = 13 c = 3.60 Now that the center, the a, b and c values, and the vertices have been determined the foci/focus needs to be determined. The formula to find the foci/focus is (h ± c, k). o ( , -1)= (0.6, -1) ( , -1)= (-6.6, -1) The last thing that has to be determined the asymptote. The equation of asymptote is y = k ± b (x h). All you have to do is input all the correct values into the equation. a o y = (x + 3) 2 y = 1 3 (x + 3) 2
4 Here is how you graph the equation and you can see that when it is graphed is it a horizontal transverse axis equation Example Problem #2 9x y 2 = 144 Find all the components of the hyperbola equation. The first step is to get the equation above in standard form. In order to do that you have to divide the whole equation by 144 because all hyperbolas always equal 1. When putting an equation in standard form make sure that the negative is always on the inside. o 16y 2 9x 2 = y 2 9 x2 16 = 1
5 Now that the given equation is in standard form you can start finding all the components of the equation. You need to determine if it is a vertical or horizontal transverse axis. o It is a vertical transverse axis because y comes first. First, the center needs to be determined. Looking at the equation there isn t a coefficient in front of x or y so, that means that the center is o Center: (0,0) Now that it is determined that the equation is a vertical transverse and you have the center. The a and b values need to be determined. In the original equation the a and b values are squared so you need to unsquared to given a and b value. o 9 = 3 16 = 4 A value =3 B value=4 Next the vertices need to be determined. The formula for vertices is (h, k ± a). o (0, 0+3)= (0,3) (0, 0-3)= (0,-3) Now that the center and vertices are determined. The c value has to be determined in order to find the foci/focus. The formula to find the c value is c 2 = a 2 + b 2 o c 2 = c 2 = c 2 = 25 c 2 = 25 c = 5 Now that the c value is determined you can find the foci/focus. The formula to find the foci/focus is (h, k ± c) o (0,0+5) =(0, 5) (0, 0-5) = (0,-5) The last thing to be determined is the asymptote. The equation for asymptote isy = k ± a (x h). All you have to do is input all the information into the equation. b o y = (x 0) 4 y = 0 3 (x 0) 4
6 Here is how you graph the equation and you can see that it is a vertical transverse axis equation.
7 Practice Problems Practice Problem #1 Practice Problem #2 6(x 3) 2 4(y + 1) 2 = 96 Find all the components of the hyperbola. y 2 25 x2 144 = 1 Find all the components of the hyperbola.
8 Practice Problems Answer Key PRACTICE PROBLEM #1 ANSWER: HORIZONTAL TRANSVERSE AXIS A=4 B=4.9 C=6.32 CENTER: (3, -1) VERTICES: (7, -1) AND (-1, -1) FOCI: (9.32, -1) AND (-3.32, -1) ASMYPTOTE: y = (x + 3) y = (x + 3) 4 PRACTICE PROBLEM #2 ANSWER: VERTICAL TRANSVERSE AXIS A=5 B=12 C=13 CENTER: (0, 0) VERTICES: (0, 5) AND (0, -5) FOCI: (0, 13) AND (0, -13) ASMYPTOTE: y = 5 12 x y = 5 12 x 4
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