How do you identity if the equation #(x-1)^2-9(y-4)^2=36# is a parabola, circle, ellipse, or hyperbola and how do you graph it?

1 Answer
Oct 25, 2016

Answer:

It is a hyperbola.
Its center is #(1, 4)#
Its vertices are :#(7, 4) ; (-5, 4)#
Its foci #(1+-2sqrt10, 4)#
Asymptots; #y=+-2/3x+10/3#

Explanation:

Given -

#(x-1)2-9(y-4)^2=36#

It is a hyperbola.

We shall write it in the standard form -

#(x-1)/36-(9(y-4)^2)/36=36/36#
#(x-1)/36-(y-4)/4=1#

It is in the form

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

Its center is #(1, 4)#

Since the #(x-1)/36#term is positive it opens right and left

Its vertices are :

#(x+a, y)#
#a^2=36#
#a=6#
#(1+6, 4)#
#(7, 4)#
#(x-a, y)#
#(1-6, 4)#
#(-5, 4)#

Its foci are -

#f=sqrt(a^2+b^2)#
#f=sqrt(36+4)=sqrt40#
#f=sqrt40=2sqrt10#

#(1+-2sqrt10, 4)#

Asymptots

#y-y_1=m(x-x_1)#

#m=#Rise/run#(Deltay)/(Deltax)=4/6=2/3#

#y-4=2/3(x-1)#

#y=2/3x-2/3+4#

#y=+-2/3x+10/3#

Refer the graph also