What is the equation of a hyperbola with vertices (3, -1) and (3, -9) and co-vertices (-6. -5) and (12, -5).

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baxthum8 | High School Teacher | (Level 3) Associate Educator

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From the information we're given we can tell that this is a vertical hyperbola, because the vertices have the same x-coordinates, 3, but different y-coordinates.  This makes them up and down from each other.  The standard equation for a vertical hyperbola is:

`(y-k)^2/a^2 - (x - h)^2/b^2 = 1` Next we'll find the center, which is represented by (h, k), as that is always half way between the vertices.  The center will be (3, -5).  Next, we'll find "a".  "a" represents the distance from the center to the vertex.  So, we want the distance from (3, -5) to either (3, -1) or (3, -9).  This distance would be 4.  Therefore a = 4.

Next we'll find "b" which is the distance from the center, (3, -5) to either of the co-vertices, (-6, -5) or (12, -5).  To find this we can use the distance formula which is:

`d =sqrt((x_(2)-x_(1))^2 + (y_(2)-y_(1))^2)`

So `d =sqrt((3-12)^2 + (-5 - (-5))^2)`

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`d =sqrt(81+0)` = 9

so "b" = 9.

Now, we can substitue all of our values into our standard equation to look like:

`(y +5)^2/4^2 - (x - 3)^2/9^2 = 1`

Simplified gives us:

`(y+5)^2/16 - (x - 3)^2/81 = 1`  

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