Question

How do you find the coordinates of the center, foci, the length of the major and minor axis given `16x^2+25y^2+32x-150y=159`?

Answer 1

The center is `=(-1,3)`
The foci are `(2,3)` and `(-4,3)`
The length of the major axis is `=10`
and the length of the minor axis is `=8`

Explanation:

Rewriting the equation
`16(x^2+2x)+25(y^2-6y)=159`
`16(x^2+2x+1)+25(y^2-6y+9)=159+16+225`
`16(x+1)^2+25(y-3)^2=400`
`(x+1)^2/25+(y-3)^2/16=1`
Comparing this to the stadard form of the ellipse
`(x-x_0)^2/a^2+(y-y_0)^2/b^2=1`
The center is `(x_0,y_0)=(-1,3)`
To calculate the foci, we need `c=sqrt(a^2-b^2)`
`c=sqrt(25-16)=3`
And the foci are `(x_0+-c,y_0)`
This lead us to F`(2,3)` and F'`(-4,3)`
The length of the major axis is `=10`
and the length of the minor axis is `=8`

graph{(x+1)^2/25+(y-3)^2/16=1 [-11.625, 10.875, -2.865, 8.385]}