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

1 Answer
Jul 17, 2018

Answer:

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=img.3...645168.646566.0.647131.6.6.0.0.0.0.61.309.6.6.0....0...1c.1.64.img..0.0.0....0.t115xwwTQyw#imgrc=yLo8-JqU_yuQ_M: