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

Jul 17, 2018

Hyperbola with transverse axis vertical

$\text{ vertex" =(0, +-2); " asymptotes: } y = \sqrt{2} x$

$\text{ foci} = \left(0 , \pm \sqrt{6}\right)$

#### Explanation:

Given: ${y}^{2} / 4 - {x}^{2} / 2 = 1$

This is a hyperbola (negative sign) with transverse axis vertical since the larger denominator is below the $y$:

${\left(y - k\right)}^{2} / {a}^{2} - {\left(x - h\right)}^{2} / {b}^{2} = 1$

where $\text{center" = (h, k); " vertex} = \left(h , k \pm a\right)$

where c = sqrt(a^2 + b^2); " foci" = (h, k+-c)

asymptotes: $y = \frac{a}{b} \left(x - h\right) + k$

${y}^{2} / {2}^{2} - {x}^{2} / {\left(\sqrt{2}\right)}^{2} = 1 \implies a = 2 , \text{ } b = \sqrt{2}$

where $\text{center" = (0, 0); " vertex} = \left(0 , \pm 2\right)$

where c = sqrt(4 + 2) = sqrt(6); " foci" = (0,+-sqrt(6))

asymptotes: $y = \frac{2}{\sqrt{2}} x \implies y = 2 \frac{\sqrt{2}}{2} x \implies y = \sqrt{2} x$

To graph manually:

Create a dashed box $\pm b$ horizontally from the center

and $\pm a$ vertically from the center. Sketch your asymptotes as dashed lines.

Place points at the vertices and foci locations. Sketch curve between the asymptotes: