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

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Answer 1 · 2018-07-22T00:22:39

See below

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#