How do you find the domain and range of #2y^2-x^2 = 8#?

1 Answer
Oct 21, 2017

Answer:

Write in the standard form of hyperbola #(y-k)^2/a^2-(x-h)^2/b^2 = 1#
The range will be #y >= k+ a# and #y <= k -a#
The domain is #x in RR#

Explanation:

Divide both sides if the given equation by 8:

#y^2/2^2-x^2/(2sqrt2)^2 = 1#

Please observe that #k = 0#, therefore, the range is:

#y >= 2# and y <= -2

The domain is #x in RR#