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

1 Answer
Jul 22, 2018

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