Vicky Shi 8-4
Math – Study Guide
All About the Equations ofHyperbolas!!>___> ‘’’
Equations of Hyperbolas Centered at the Origin:
(x^2 / a^2) – (y^2 / b^2) = 1
When x^2 is over a^2, the hyperbola will open left and right, like this –
(y^2 / a^2) – (x^2 / b^2) = 1
When y^2 is over a^2, the hyperbola will open up and down, like this –
Equations of Translated Hyperbolas:
[(x-h)^2 / a^2] – [(y-k)^2 / b^2] = 1
When (x-h)^2 is over a^2, the hyperbola will open left and right, like this –
[(y-k)^2 / a^2] – [(x-h)^2 / b^2] = 1
When (y-k)^2 is over a^2, the hyperbola will open up and down, like this –
The “h” will always go with “x” and the “k” will always go with “y,” to matter which of the pair is over “a.” The color scheme will in the equations above will help you later in these study guide [I hope…].
**These are how equations for hyperbolas look like, so memorize them!!**
Identifying a Hyperbola!! YAY!! (.__.)n
A hyperbola has a lot of important things that you need to know the names of and how to find them. They are:
-the CENTER (h,k)
-the TRANSVERSE AXIS
-the VERTICES or the “a” VALUES
-the CONJUGATE AXIS
-the CO-VERTICESor the “b” VALUES
-the FOCI or the “c” VALUES
-the ASYMPTOTES (ignore the blue points on the second graph for this one)
I tried getting the colors of the test to correspond as closely to the graphs as I could, so sorry if they’re not exact…
**Use these graphs as a reference to understand the explanation below!!**
The Center:The center of a hyperbola is defined by (h,k). The “x” value is equal to the “h” value, same for “y” and “k.”If the equation of a hyperbola look this (x^2 / a^2) – (y^2 / b^2) = 1 or (y^2 / a^2) – (x^2 / b^2) = 1, then the center is and always will be the origin, (0,0). But if the hyperbola is translated, or the equations look like these [(x-h)^2 / a^2] – [(y-k)^2 / b^2] = 1 or [(y-k)^2 / a^2] – [(x-h)^2 / b^2] = 1, the center is where the two asymptotes of the hyperbola intersect, or the “h” and “k” value of the equation, which will probably be given to you.(Important info about origin hyperbolas are highlighted in yellow, translated hyperbolas in cyan.)
The transverse axis, “a” & the vertices:These three items ALWAYS GO TOGETHER. Remember that! The vertices are the points that touch the hyperbola and they have the same “x” or “y” value [depending on how the hyperbola opens] as the center. The vertices are found by square rooting a^2 in the hyperbola equations, in other words, the positive and negative value of “a.” If the hyperbola is centered at the origin and opens left and right, the vertices are (a,0) and (-a,0)and if the hyperbola opens up and down, the vertices are (0,a) and (0,-a). If the hyperbola is translated, then to find the vertices, you add and subtract the “a” value from the center; for translated hyperbolas that opens left and right, (h+a,k) and (h-a,k), and for translated hyperbolas that open up and down, (h,k+a) and (h,k-a). The transverse axis is just the distance between the two vertices. (Important info about the three items listed above is in red text.)
The conjugate axis, “b” & the co-vertices: These three items also ALWAYS GO TOGETHER.The co-vertices don’t touch the hyperbola, but they still have a same “x” or “y” value [depending on how the hyperbola opens] as the center. The co-vertices are found by square rooting b^2 in the hyperbola equations; the positive and negative value of “b.” If the hyperbola is centered at the origin and opens left and right, the co-vertices are (0,b) and (0,-b) and if the hyperbola opens up ans down, the vertices are (b,0) and (-b,0). If the hyperbola is translated, then to find the co-vertices, you add and subtract the “b” value from the center; for translated hyperbolas that opens left and right, (h,k+b) and (h,k-b), and for translated hyperbolas that open up and down, (h+b,k) and (h-b,k). The conjugate axis is just the distance between the two co-vertices. (Important info about the three items listed above is in green text.)
The foci & “c”:To find the foci of hyperbolas, use the Pythagorean Theorem or a^2 + b^2 = c^2, or c = √a^2 + b^2.Just take the two bottom numbers from the given hyperbola equation, add them, and then take the square root of the sum of the numbers. If the hyperbola is centered at the origin and opens left or right, the foci are (c,0) and (-c,0) and if the hyperbola opens up and down, the vertices are (0,c) and (0,-c).If the hyperbola is translated and opens left and right, the foci are (h+c,k) and (h-c,k) and if the hyperbola opens up and down, the foci are (h,k+c) and (h,k-c).Also, using focus-focus graph paper, the distance from focus1 to a point on the first half of the hyperbola to focus2 should be the same from the focus2 to a point on the second half of the hyperbola to focus1. (Important info about the two items listed above is in purple text.)
The asymptotes: The equation for the asymptotes of a hyperbola are y = (b/a)x + b and y = -(b/a)x + b if the hyperbola opens left and right, and y = (a/b)x + b and y = -(a/b)x +b if the hyperbola opens up and down. If you look at the second graph from above, the asymptotes go through the corners of the blue-dashed box. This will happen for all asymptote equations related to hyperbolas, although it won’t be the same box as the one shown above. The asymptotes are the lines the hyperbola reaches for but never touch.(Important info about the asymptotes is in blue text.)
Practice, Practice,Practice!! =D(arrghh... >__>)
Find the center, vertices, co-vertices, foci and asymptotes for these to graphs, as well as the equations for these hyperbolas.
- 2.
Graph these equations and find the center, vertices, co-vertices, foci and asymptotes:
- (y^2/9) – (x^2/25) =1
- [(x-2)^2/25 – (y+2)^2/36] = 1
- x^2/4 – (y-3)^2/25 = 1
Complete the Square with Hyperbolas!! =______= ‘’’
Hyperbolic equations can be found looking like this:
4x^2 – 9y^2 – 8x + 54y = 113
In order to translate that into the standard form of a hyperbolic equation, use the “complete the square” technique.
First, rearrange the equation so the “x’s” and “y’s” are together in groups. Using parentheses might help.
4x^2 – 9y^2 – 8x + 54y = 113to (4x^2 – 8x) + (-9y^2 +54y) = 113
Next, extract the “a” term, as in ax^2 + bx – c, from within the parentheses.
(4x^2 – 8x) + (-9y^2 +54y) = 113 to 4(x^2 – 2x) - 9(y^2 - 6x) = 113
**Notice that because the “a” term in the second part of the equation is negative, the signs flip when it’s taken out**
The next step is to complete the square inside the parentheses. To find to “c” value (in ax^2 + bx +c), take the “b,” divide it by 2, and square it; c = (b/2)^2. Don’t for get to add these new values to the other side of the equation as well. And DON’T FORGET to multiply the “c” value by the coefficient on the outside of the parentheses.
4(x^2 – 2x) - 9(y^2 - 6x) = 113
Becomes 4(x^2 – 2x + 1) – 9(y^2 – 6x + 9) = 113 + 4 – 81
Then just simplify this equation.
4(x^2 – 2x + 1) – 9(y^2 – 6x + 9) = 113 + 4 – 81
Becomes 4(x-1)^2 – 9(y – 3)^2 = 36
Next step is to divide so the equation ends up equaled to 1.
4(x-1)^2 – 9(y – 3)^2 = 36
Becomes [(x-1)^2 / 9 – (y – 3)^2 / 4] = 1 which is in standard hyperbola equation form.
Practice, Practice, Practice!! =D Part 2 (*sigh* u.u)
Write these equations in standard hyperbola equation form.
- 25y^2 – 16x^2 + 64x – 50y = 439
- 4y^2 – 36x^2 – 72x + 8y = 176
- 16x^2 - 25y^2 – 32x + 100y = 484
ANSWERS!! Can You Find Them??+ w +(MUAHAHAHA!! No cheating!!)
- y^2/4 – x^2/9 = 1
Center (0,0)
Vertices (0,2) and (0,-2)
Co-Vertices (3,0) and (-3,0)
Foci (0, √13) and (0,-√13)
Asymptotes y = 2/3x and y = -2/3x
- [(x-1)^2/4 – (y+1)^2/9] = 1
Center (1,-1)
Vertices (-1,-1) and (3,-1)
Co-Vertices (1,2) and (1,-4)
Foci (1+√13,-1) and (1-√13,-1)
Asymptotes y = 3/2x – 2.5 and y = -3/2x + 0.5
- Sorry, I can’t graph it…
Center (0,0)
Vertices (0,3) and (0,-3)
Co-Vertices (5,0) and (-5,0)
Foci (0,√34) and (0,-√34)
Asymptotes y = 3/5x and y = -3/5x
- Sorry, I can’t graph it…
Center (2,-2)
Vertices (7,-2) and (-3,-2)
Co-Vertices (2,4) and (2,-8)
Foci (2+√61,-2) and (2-√61,-2)
Asymptotes y = 6/5x – 4.4 and y = -6/5x + 0.4
- Sorry, I can’t graph it…
Center (0,3)
Vertices (2,3) and (-2,3)
Co-Vertices (0,8) and (0,-2)
Foci (√29,0) and (-√29,0)
Asymptotes y = 5/2x +3 and y= -5/2x + 3
- [(y-1)^2/16 – (x-2)^2/25] = 1
- [(y+1)^2/36 – (x+1)^2/4] = 1
- [(x-1)^2/25 – (y-2)^2/16] = 1
**In order to see the answers, highlight this page and the page above this one black, like this**