Question

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

Answer 1

Hyperbola with transverse axis vertical

`" vertex" =(0, +-2); " asymptotes: " y = sqrt(2)x`

`" foci" = (0, +-sqrt(6))`

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`:

`(y-k)^2/a^2 - (x-h)^2/b^2 = 1`

where `"center" = (h, k); " vertex" =(h, k+-a)`

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

asymptotes: `y = a/b (x - h)+k`

`y^2/2^2 - x^2/(sqrt(2))^2 = 1 => a = 2, " "b = sqrt(2)`

where `"center" = (0, 0); " vertex" =(0, +-2)`

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

asymptotes: `y = 2/sqrt(2) x => y = 2sqrt(2)/2 x => y = sqrt(2)x`

To graph manually:

Create a dashed box `+-b` horizontally from the center

and `+-a` vertically from the center. Sketch your asymptotes as dashed lines.

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

![https://www.google.com/search?tbm=isch&sa=1&ei=vm5NW92CGYqsjwSmoLPQCQ&q=graph+foci+of+hyperbola+with+transverse+axis+vertical&oq=graph+foci+of+hyperbola+with+transverse+axis+vertical&gs_l=