How do you find the vertices and foci of #36x^2-10y^2=360#?

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Answer 1 · 2016-11-17T07:05:39

The vertices are #=(sqrt10,0)# and #=(-sqrt10,0)#
The foci are F#=(sqrt46,0)# and F'#=(-sqrt46,0)#
The equations of the asymptotes are

#y=6/sqrt10x# and #y=-6/sqrt10x#

Let's rewrite the equation

#36x^2-10y^2=360#

Divide by 360

#x^2/10-y^2/36=1#

This is the equation of a right left hyperbola

#x^2/a^2-y^2/b^2=1#

The center is #=(0,0)#

The vertices are #(+-a,0)#
#=(+-sqrt10,0)#

To calculate the foci, we need #c=+-sqrt(a^2+b^2)#

#c=+-sqrt(10+36)=+-sqrt46#

The foci are F#=(sqrt46,0)# and F'#=(-sqrt46,0)#

The slope of the asymptotes are #=+-b/a#

#=+-6/sqrt10#

The equations of the asymptotes are

#y=6/sqrt10x# and #y=-6/sqrt10x#

graph{(x^2/10-y^2/36-1)(y-6/sqrt10x)(y+6/sqrt10x)=0 [-28.87, 28.87, -14.44, 14.44]}