Dan P.

asked • 08/02/19

Hyperbola: a² > b²

In precalculus, the standard equation of a hyperbola with center at coordinates (0,0) is x² / a² - y² / b² = 1.

Is it true that a² > b².

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Mark M. answered • 08/02/19

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No. For a hyperbola, it is possible that a2 < b2. The length of the transverse axis is 2a and the length of the conjugate axis is 2b. The length of the conjugate axis is not always larger than the length of the transverse axis.


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Dan P.

Thank you. Can you give an example sir.
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08/04/19

Mark M.

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16x^2 - 9y^2 = 144 is a hyperbola. The equation can also be written as x^2/9 - y^2/16 = 1.
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08/04/19

Jayson K. answered • 08/02/19

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No, unlike the ellipse where a2 > b2, that is not necessarily true for a hyperbola. In fact there will be times when a2 = b2 or a2 < b2. Only the ellipse holds this property.


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Mr. K

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Dan P.

Thank you. Can you give an example sir?
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08/04/19

Jayson K.

Sure, as you can see above there already are a few examples. The key to realize here though is that with hyperbolas, your formulas are either x^2/a^2 - y^2/b^2 = 1 or y^2/a^2 - x^2/b^2 = 1 The key here is to notice that with the equation(s) for a hyperbola, the a^2 is always the front number and the b^2 is always the back number. So if you have x^2/4 - y^2/9 = 1 here a^2 = 4 and b^2 = 9; x^2/16 - y^2/4 =1 here a^2 = 16, and b^2 = 4 x^2 - y^2 = 1 here a^2 = 1 = b^2 As you can see, the value of a^2 and b^2 is not going to matter which one is greater, the only thing that matters is which guy is in front. With an ellipse on the other hand, x^2/a^2 + y^2/b^2 = 1 or x^2/b^2 + y^2/a^2 = 1 to find a^2 you are looking for the larger of the two numbers, as that would be the only way to distinguish between the two numbers. So, x^2/4 + y^2/9 = 1 here a^2 = 9 and b^2 = 4; x^2/16 + y^2/4 =1 here a^2 = 16, and b^2 = 4 x^2 + y^2 = 1 though, a^2 = 1 = b^2 this is NOT an ellipse, it's classified as a circle! Hope this helps Mr. K
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08/04/19

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