How do you graph #x^2/16-y^2/4=1# and identify the foci and asympototes?

1 Answer
Jul 27, 2016

Answer:

The eqns. of Asymptotes are #y=+-1/2x#.

Focii are #S(ae,0)=S(2sqrt5,0) and S'(-ae.0)=S'(-2sqrt5,0)#.

Explanation:

I will help find the Focii and Asymptotes without graphing the Hyperbola #(1) : x^2/16-y^2/4=1#.

The Std. Eqn. of the Hyperbola is given by, #(2) :x^2/a^2-y^2/b^2=1#. Its Asymptotes are easily found by just replacing #1# [from the Std. Eqn.] by #0#. In other words, the Asymptotes are,

#x^2/a^2-y^2/b^2=0, or, y=+-b/ax#.

In our case, the eqns. of Asymptotes are #y=+-1/2x#.

For Focii, we have to find the Eccentricity #e# of the Hyperbola, defined by, #b^2=a^2(e^2-1)#

#:.4=16(e^2-1)rArr4/16+1=e^2=20/16=5/4rArre=sqrt5/2#.

Hence, Focii are #S(ae,0)=S(4*sqrt5/2,0)=S(2sqrt5,0)# and

#S'(-ae.0)=S'(-2sqrt5,0)#.