Question

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

Answer 1

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)`.