How do you find all the critical points to graph #4x^2-9y^2=36# including vertices, foci and asymptotes?

1 Answer

Answer:

Vertices #(+-3, 0)#
Focii #(+-sqrt13, 0)#
Asymptotes equations: #y=+-2/3x#

Explanation:

The equation #4x^2-9y^2=36# is a hyperbola

The standard form of the equation of a hyperbola is

#(x-h)^2/a^2-(y-k)^2/b^2=1#

For the given #4x^2-9y^2=36#

We divide both sides of the equation by #36#

#(4x^2-9y^2)/36=36/36#

#x^2/9-y^2/4=1#

it follows that

#(x-0)^2/(3^2)-(y-0)^2/(2^2)=1#

and by inspection and by comparing to the form above,

#(x-h)^2/a^2-(y-k)^2/b^2=1#

we have

#a=3# and #b=2# and
#c=sqrt(a^2+b^2)=sqrt(9+4)=sqrt13#
#c=sqrt13#

and center #(h, k)=(0, 0)#

Center #(h, k)=(0, 0)#
Vertices at #(h+a, k)= (+3, 0)#
and #(h-a, k)=(-3, 0)#

Focii at #(h+c, k)=(+sqrt13, 0)=(3.6, 0)#
and #(h-c, k)=(-sqrt13, 0)=(-3.6, 0)#

The equations of the asymptotes are given by the following formulas

#y=+-b/a(x-h)+k#

therefore,

#y=+-2/3(x-0)+0#

we have

#y=2/3x# and #y=-2/3x#

Kindly see the graphs of the hyperbola #4x^2-9y^2=36# colored red and asymptotes #y=2/3x# and #y=-2/3x# colored blue. You can also see the colored orange dots (Focii) and colored green dots (Vertices).

Desmos

God bless... I hope the explanation is useful.