# Find the equation of the hyperbola. Center(2,2), Vertex(2,4), and goes through the point (2 +√10, 5) please show how to solve the problem

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### 1 Answer

You should write the standard equation of hyperbola centered at `(h,k)` such that:

`(x - h)^2/a^2- (y - k)^2/b^2 = 1`

Since the problem provides the coordinates of the center, `h = 2` and `k = 2` , you may substitute the coordinates in equation above, such that:

`(x - 2)^2/a^2 - (y - 2)^2/b^2 = 1`

The problem provides the coordinates of the vertex of hyperbola, hence, you may find a, since the distance from the center to each vertex of hyperbola yields a.

`a = sqrt((2-2)^2 + (4-2)^2) => a = 2`

You may complete the equation of hyperbola substituting 2 for a such that:

`(x - 2)^2/2^2 - (y - 2)^2/b^2 = 1`

You need to find b, hence, you should use the last information provided by the problem that the point `(2+sqrt10,5)` lies on hyperbola such that:

`(2 + sqrt10- 2)^2/2^2 - (5 - 2)^2/b^2 = 1`

`10/4 - 9/b^2 = 1 => 10b^2 - 36 = 4b^2 => 10b^2 - 4b^2 = 36`

`6b^2= 36 => b^2 = 6`

Now, the equation of hyperbola is fully determined:

`(x - 2)^2/2^2 - (y - 2)^2/(sqrt6)^2 = 1`

**Hence, evaluating the equation of hyperbola, under the given conditions, yields `(x - 2)^2/2^2 - (y - 2)^2/(sqrt6)^2 = 1` .**