How do you find all the critical points to graph #36x^2 - 100y^2 - 72x + 400y = 3964# including vertices, foci and asymptotes?

1 Answer
Sep 28, 2017

Answer:

Vertices are #(11,2)# and #(-9,2)#, foci are (#1±2sqrt(34),2#) and asymptotes are #y=3/5x+7/5# and #y= -3/5x+13/5#.

Explanation:

The graph is for a hyperbola.
The standard form of equation for a hyperbola is
#x^2/(a^2) - y^2/(b^2) = ±1#. #(a>0, b>0)#
What we have to do is to deform the fomula to this form.

[step1] Let's complete the square.
#36x^2 -100y^2 -72x +400y =3964#
#36(x^2-2x) -100(y^2-4y)=3964#
#36(x-1)^2-100(y-2)^2=3964+36×1-100×4#
#36(x-1)^2-100(y-2)^2=3600#

[step2] divide both side by #3600# to make a #1# on the right side.
#(x-1)^2/(10^2) - (y-2)^2/(6^2) = 1#

The hyperbola has vertices and foci on on the line #y=2#.
Substituting #y=2# to the fomula yields #x=11,-9#
Therefore, the vertices are #(11,2)# and #(-9,2)#.

The foci of #x^2/(a^2) - y^2/(b^2) = 1# are #(±sqrt(a^2+b^2),0)# .
If the fomula were #x^2/(10^2) - y^2/(6^2) = 1#, the foci would be
(#±sqrt(6^2+10^2),0#) = (#±2sqrt(34),0#). Translate this #1# to the right and #2# to the up and the foci are (#1±2sqrt(34),2#).

The asymptopes of #x^2/(a^2) - y^2/(b^2) = 1# are #y=±(a/b)x# and the asymptopes of #(x-1)^2/(10^2) - (y-2)^2/(6^2) = 1# are therefore
#y-2 = ±(6/10)(x-1)#
#y=3/5x+7/5# and #y= -3/5x+13/5#.