How do you find the standard form of #9x^2-16y^2-90x+32y+65=0#?

1 Answer
Apr 15, 2016

Answer:

#(x-5)^2/4^2-(y-1)^2/3^2=1#

Explanation:

#9x^2-90x=9(x-5)^2-225#.
#-19y^2+32y=-16(y-1)^2+16#.

So, the given equation becomes
#9(x-5)^2-16(y-1)^2=144#
Divide by 144.
#(x-5)^2/4^2-(y-1)^2/3^2=1#.

This equation represents a hyper bola with semi-major axis a = 4 and semi-transverse axis b = 3.

The eccentricity of the hyperbola is 1.25, using the relation #b^2=a^2(e^2-1)#

The asymptotes are #y=+-(b/a)x=+-3/4x#.