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
It is a hyperbola.
Its center is
Its vertices are :
Its foci
Asymptots;
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
Since the
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#