How do you write #x^2 - 8y^2 - 14x -15 = 0# in standard form?
1 Answer
May 21, 2015
This is a hyperbola with a horizontal transverse axis so its standard form is
Given
#x^2-14x+49 - 8y^2 = 15 +49#
#(x-7)^2-8y^2 = 64#
#(x-7)^2/8^2 - y^2/(2sqrt(2))^2 = 1#
