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Asymptotes of a hyperbola are the lines that pass through center of the hyperbola. The hyperbola gets closer and closer to the asymptotes, but can never reach them. There are two different approaches you can use to find the asymptotes. Learning how to do both may help you understand the concept.
Steps
Method 1
Method 1 of 2:
Factoring
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1Write down the equation of the hyperbola in its standard form. We'll start with a simple example: a hyperbola with the center of its origin. For these hyperbolas, the standard form of the equation is x2/a2 - y2/b2 = 1 for hyperbolas that extend right and left, or y2/b2 - x2/a2 = 1 for hyperbolas that extend up and down. Remember, x and y are variables, while a and b are constants (ordinary numbers).[1]
- Example 1: x2/9 - y2/16 = 1
- Some textbooks and teachers switch the position of a and b in these equations. Follow the equation closely so you understand what's going on. If you just memorize the equations you won't be prepared when you see a different notation.
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2Set the equation equal to zero instead of one. This new equation represents both asymptotes, though it will take a little more work to separate them.[2]
- Example 1: x2/9 - y2/16 = 0
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3Factor the new equation. Factor the left hand side of the equation into two products.[3] Refresh your memory on factoring a quadratic if you need to, or follow along while we continue Example 1:
- We'll end up with an equation in the form (__ ± __)(__ ± __) = 0.
- The first two terms need to multiply together to make x2/9, so take the square root and write it in those spaces: (x/3 ± __)(x/3 ± __) = 0
- Similarly, take the square root of y2/16 and place it in the two remaining spaces: (x/3 ± y/4)(x/3 ± y/4) = 0
- Since there are no other terms, write one plus sign and one minus sign so the other terms cancel when multiplied: (x/3 + y/4)(x/3 - y/4) = 0
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4Separate the factors and solve for y. To get the equations for the asymptotes, separate the two factors and solve in terms of y.[4]
- Example 1: Since (x/3 + y/4)(x/3 - y/4) = 0, we know x/3 + y/4 = 0 and x/3 - y/4 = 0
- Rewrite x/3 + y/4 = 0 → y/4 = - x/3 → y = - 4x/3
- Rewrite x/3 - y/4 = 0 → - y/4 = - x/3 → y = 4x/3
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5Try the same process with a harder equation. We've just found the asymptotes for a hyperbola centered at the origin. A hyperbola centered at (h,k) has an equation in the form (x - h)2/a2 - (y - k)2/b2 = 1, or in the form (y - k)2/b2 - (x - h)2/a2 = 1. You can solve these with exactly the same factoring method described above. Just leave the (x - h) and (y - k) terms intact until the last step.
- Example 2: (x - 3)2/4 - (y + 1)2/25 = 1
- Set this equal to 0 and factor to get:
- ((x - 3)/2 + (y + 1)/5)((x - 3)/2 - (y + 1)/5) = 0
- Separate each factor and solve to find the equations of the asymptotes:
- (x - 3)/2 + (y + 1)/5 = 0 → y = -5/2x + 13/2
- ((x - 3)/2 - (y + 1)/5) = 0 → y = 5/2x - 17/2
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Method 2
Method 2 of 2:
Solving for Y
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1Write down the hyperbola equation with the y2 term on the left side. This method is useful if you have an equation that's in general quadratic form. Even if it's in standard form for hyperbolas, this approach can give you some insight into the nature of asymptotes. Rearrange the equation so the y2 or (y - k)2 term is on one side to get started.[5]
- Example 3: (y + 2)2/16 - (x + 3)2/4 = 1
- Add the x term to both sides, then multiply each side by 16:
- (y + 2)2 = 16(1 + (x + 3)2/4)
- Simplify:
- (y + 2)2 = 16 + 4(x + 3)2
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2Take the square root of each side. Take the square root, but don't try to simplify the right hand side yet. Remember, when you take the square root, there are two possible solutions: a positive and a negative. (For example, -2 * -2 = 4, so √4 can be equal to -2 as well as 2.) Use the "+ or -" sign ± to keep track of both solutions.[6]
- √((y + 2)2) = √(16 + 4(x + 3)2)
- (y+2) = ± √(16 + 4(x + 3)2)
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3Review the definition of an asymptote. It's important that you understand this before you continue to the next step. The asymptote of a hyperbola is a line that the hyperbola gets closer and closer to as x increases. X can never actually reach the asymptote, but if we follow the hyperbola for larger and larger values of x, we'll get closer and closer to the asymptote.[7]
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4Adjust the equation for large values of x. Since we're trying to find the asymptote equation now, we only care about x for very large values ("approaching infinity"). This lets us ignore certain constants in the equation, because they contribute such a small part relative to the x term. Once x is at 99 billion (for example), adding three is so small we can ignore it.
- In the equation (y+2) = ± √(16 + 4(x + 3)2), as x approaches infinity, the 16 becomes irrelevant.
- (y+2) = approximately ± √(4(x + 3)2) for large values of x
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5Solve for y to find the two asymptote equations. Now that we've got rid of the constant, we can simplify the square root. Solve in terms of y to get the answer. Remember to split the ± symbol into two separate equations, one with + and one with -.[8]
- y + 2 = ±√(4(x+3)^2)
- y + 2 = ±2(x+3)
- y + 2 = 2x + 6 and y + 2 = -2x - 6
- y = 2x + 4 and y = -2x - 8
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Community Q&A
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QuestionHow do I find the equation of asymptotes of hyperbolis 16x2-9y2=144?
Community AnswerDivide both sides of the equation by 144 to get 1 on the right hand => the equation will be x^2/9 + y^2/16 =1 => a=3 and b=4 so the equation of asymptote will be y = - b/a x and y= b/a x so y= - 4/3*x and y = 4/3*x. -
QuestionHow do I solve the equation of f(x)+3?
Community AnswerThat's not an equation. You would need two sides of an equation, divided by an equal sign (=), in order to solve it. -
QuestionHow to find the asymptotes of a rectangular hyperbola whose equation is x squared - y squared = 1?Community AnswerIn this case a and b are both equal to 1, and the asymptotes are y = x and y = -x.
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Warnings
- Beware of putting equations always in standard form.⧼thumbs_response⧽
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References
- ↑ https://flexbooks.ck12.org/cbook/ck-12-math-analysis-concepts/section/6.7/primary/lesson/hyperbolas-and-asymptotes-mat-aly/
- ↑ https://www.cuemath.com/jee/asymptotes-hyperbolas/
- ↑ https://flexbooks.ck12.org/cbook/ck-12-math-analysis-concepts/section/6.7/primary/lesson/hyperbolas-and-asymptotes-mat-aly/
- ↑ https://www.cuemath.com/jee/asymptotes-hyperbolas/
- ↑ https://courses.lumenlearning.com/waymakercollegealgebra/chapter/equations-of-hyperbolas/
- ↑ https://math.libretexts.org/Bookshelves/Algebra/Map%3A_College_Algebra_(OpenStax)/08%3A_Analytic_Geometry/8.03%3A_The_Hyperbola
- ↑ https://math.libretexts.org/Bookshelves/Algebra/Map%3A_College_Algebra_(OpenStax)/08%3A_Analytic_Geometry/8.03%3A_The_Hyperbola
- ↑ https://courses.lumenlearning.com/waymakercollegealgebra/chapter/equations-of-hyperbolas/
About This Article
Article SummaryX
To find the equations of the asymptotes of a hyperbola, start by writing down the equation in standard form, but setting it equal to 0 instead of 1. Then, factor the left side of the equation into 2 products, set each equal to 0, and solve them both for “Y” to get the equations for the asymptotes. Alternatively, you can rearrange the equation with the Y^2 term on the left side, take the square root of both sides, then solve for “Y.” Just remember to split your answer into 2 separate equations, one with a plus sign and the other with a minus sign. To work through examples of how to find the equations of the asymptotes of a hyperbola using both these methods, read on!
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