Question

The equation is x²-y²+6x+2y=10. What is the center,vertices,foci length of Latus Rectum, asymptotes and the graph of hyperbola?

Answer 1

See below

Explanation:

Let us start by putting this equation in its standard form:
`x²-y²+6x+2y=10`

`x^2+6x-y^2+2y=10`

`(x+3)^2-(y-1)^2=10+9-1`

`(x+3)^2/(18)-(y-1)^2/(18)=1`

`vertex: (-3, 1)`

`a= sqrt18=3sqrt2`
`b= sqrt18= 3sqrt2`

`c= sqrt(a^2+b^2)`

`c= sqrt36= 6`

To get vertices, add and subtract `a` to the `x` coordinates of the vertex:
`Vertices: (-3+3sqrt2, 1) and (-3-3sqrt2, 1)`

To get focii, add and subtract the `c` to the `x` coordinates of the vertex:
`Focii: (3, 1) and (-9, 1)`

Length of the latus rectum:
`L=2b^2/a= (2*18)/(3sqrt2)=6sqrt2`
Now graph these points using estimations

For the asymptotes, simply go `b/a` in both directions from the center:
`y=x+4`
`y=-x-2`