# Find the equation of the ellipse with a center:(0,0); vertex:(5,0) and containes (sqrt of 15,2).

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### 1 Answer

The given are:

Center: `(0,0)`

Vertex: `(5,0)`

and a point along the ellipse : `(sqrt15 , 2)`

Since center and vertex have the same y-coordinates, then the major axis of the ellipse is horizontal.

For ellipse with horizontal major axis, the standard equation is:

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

where `(h,k)` is the center,

`a` is the distance between vertex and center or half length of major axis , and

`b` is the half length of the minor axis

Note that `a` and `b` must be solved to be able to determine the equation of the ellipse.

To compute for `a` , subtract the x-coordinate of the center from the vertex.

`a = 5-0 = 5`

To solve for b, substitute `a=5` , center`(0,0)` and point`(sqrt15,2)` to the formula of ellipse.

`(sqrt15-0)^2/5^2 + (2-0)^2/b^2=1` `4/b^2= 2/5`

`15/25+4/b^2 = 1` `4* 5/2 = b^2`

`3/5 + 4/b^2 =1` `10 = b^2`

`4/b^2 = 1-3/5 ` `+-sqrt10 = b`

Since `b` is a length, take the positive value. So, `b=sqrt10` .

Then, substitute `a=5` , `b=sqrt10` and center`(0,0)` to the formula of ellipse.

`(x-0)^2/5^2 + (y-0)^2/(sqrt10)^2 = 1`

`x^2/25 + y^2/10 = 1`

**Hence, the equtaion of the ellipse with center `(0,0)` , vertex `(5,0)` and containing the point `(sqrt15,2)` is `x^2/25+y^2/10=1` .**

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