Hyperbolbros Emma Romigh, Jalen Cornell, Bailey Cosper, Evan Moore, Cassidy Williams, Brenna Maxwell

Vertex and Standard form Introduction Vertex/General Form Standard form

Identifying parts of a Hyperbola The hyperbola is centered on a point (h, k), which is the "center" of the hyperbola. The point on each branch closest to the center is that branch's "vertex". The vertices are some fixed distance a from the center. The line going from one vertex, through the center, and ending at the other vertex is called the "transverse" axis. The "foci" of an hyperbola are "inside" each branch, and each focus is located some fixed distance c from the center. The values of a and c will vary from one hyperbola to another, but they will be fixed values for any given hyperbola.

Using an equation to solve for c “C” is the distance from the focus to the center. It is a fixed value for any given parabola, but can vary between parabolas. To find “c” using an equation, you must be sure it is in conics form, you can do this by completing the square. Then you solve a^2 + b^2 = c^2. Example: 4x 2 – 5y x – 30y – 45 = 0. 4x x – 5y 2 – 30y = 45 4(x x ) – 5(y 2 + 6y ) = ( ) – 5( ) 4(x x + 25) – 5(y 2 + 6y + 9) = (25) – 5(9) 4(x + 5) 2 – 5(y + 3) 2 = – 45 a 2 = 25 and b 2 = 20, so a = 5 and b = 2sqrt[5]. The equation a 2 + b 2 = c 2 gives me c 2 = = 45, so c =sqrt[45] = 3sqrt[5]. ath.com/modules/hy perbola2.htm

Write an equation of a hyperbola given... V ertices and foci: Example of finding an equation of the hyperbola with the given foci and vertices. Foci: (–10, 0), (10, 0), Vertices: (–9, 0), (9, 0) 1.Find whether the transverse axis is vertical or horizontal The vertices and the foci lie on the x –axis. Therefore, the transverse axis is horizontal. 2. The foci and the vertices are equidistant from the origin. Find the origin. The center is the origin since the foci and the vertices are equidistant from the origin. 3. Find c and a, from the foci and the vertices. Since the foci are each 10 units from the center, c = 10. Similarly, the vertices are each 9 units from the center, a = 9.

Write an equation of a hyperbola given... Given vertices and foci continued: 4. Find b using the equation b 2 = c 2 – a 2. Substituting 9 for a and 10 for c in the equation b 2 = c 2 – a 2 : b 2 = 10 2 – 9 2 b 2 = 100 – 81 5.Substitute for a and b in the standard form of the hyperbola with horizontal transverse axis The transverse axis is horizontal. Thus the standard form of the hyperbola is: Substituting 9 for a and √ 19 for b:

Write an equation of a hyperbola given... Transverse axis length and focus:

Write an equation of a hyperbola given... Given vertices and asymptotes: Practice Vertices: (13,0), (-1,0) Asymptotes: y=x-6 y=-x+6+ Did you get this? Answer: (x-6)^2/49 - y^2/49 = 1

Sketching a graph from standard form -example will be provided by instructor

Standard form to complete a square 1. 9x²+36x-y²+10y+2=0 2. 9x²+36x-y²+10y= (x²+4x)-(y²-10y)= (x²+4x+4)-(y²-10y+25)= (x2+4x+4)-(y2-10y+25)= (x+2)²-(y-5)²= Make sure variables are grouped together 2.Rearrange the equation so that variables are on one side while coefficients are on the other 3.Factor out a coefficient of the x and y terms 4.Take the number in front of the first degree term, divide it by 2, then square it. This is your third number in your parentheses 5.You must now add these numbers to the other side. Remember to distribute the coefficient when moving these to the other side (in this case there is an implied -1 in front of the second set of parentheses) 6.Factor each trinomial and simplify 7.Make the right side equal 1 by dividing each term by 9 in this case 8.Simplify 2RRUagUc

Standard form to vertex form 1. First you need to get rid of the fractions by multiplying both sides by the denominators and simplifying. 2. Next you need to expand your squared terms to get binomials. 3. The next step is to distribute your coefficients to each term (Don’t forget the negatives) 4. Finally you combine like terms and set the right side equal to zero. Try this: