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

Jul 27, 2016

The eqns. of Asymptotes are $y = \pm \frac{1}{2} x$.

Focii are $S \left(a e , 0\right) = S \left(2 \sqrt{5} , 0\right) \mathmr{and} S ' \left(- a e .0\right) = S ' \left(- 2 \sqrt{5} , 0\right)$.

Explanation:

I will help find the Focii and Asymptotes without graphing the Hyperbola $\left(1\right) : {x}^{2} / 16 - {y}^{2} / 4 = 1$.

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

${x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 0 , \mathmr{and} , y = \pm \frac{b}{a} x$.

In our case, the eqns. of Asymptotes are $y = \pm \frac{1}{2} x$.

For Focii, we have to find the Eccentricity $e$ of the Hyperbola, defined by, ${b}^{2} = {a}^{2} \left({e}^{2} - 1\right)$

$\therefore 4 = 16 \left({e}^{2} - 1\right) \Rightarrow \frac{4}{16} + 1 = {e}^{2} = \frac{20}{16} = \frac{5}{4} \Rightarrow e = \frac{\sqrt{5}}{2}$.

Hence, Focii are $S \left(a e , 0\right) = S \left(4 \cdot \frac{\sqrt{5}}{2} , 0\right) = S \left(2 \sqrt{5} , 0\right)$ and

$S ' \left(- a e .0\right) = S ' \left(- 2 \sqrt{5} , 0\right)$.