Find the values of a and b that make the hyperbola (x^2/a^2) - (y^2/b^2) =1 congruent to xy =1.

Question

PLEASE NEED HELP TODAY!! This deals with conics

Answer 1

first hyperbola rotated 45 degrees about the origin can give the 2nd hyperbola.

the 2nd is rectangular has axis of symmetry x=y

the 1st has axis of symmetry y=0

asymptotes for the 1st are y =b/a and y =-b/a,

asymptotes for the 2nd are the x and y axes, y=0 and x= 0

vertices of the 1st are (a,0) and (-a,0)

vertices for the 2nd are (1,1) and (-1,-1)

distance from the vertices for the 2nd = sqr8

distance between the vertices fot the 1st 2a

for the hyperolas to be similar, make 2a = sqr8 = 2sqr2

a = sqr2 = b

set a and b = sqr2

x^2/2 - y^2/2 = 1,

or

x^2 -y^2 = 2

rotate it 45 degrees counterclockwise around the origin and you get xy = 1

branches of the 2nd hyperbola are in quadrant I and III

branches of the 1st have one branch ins quadrant II&III, and the other branch in quadrant I&IV

it might help to sketch the two hyperbolas

x^2 - y^2 = 1 is affinely equivalent to y = 1/x or xy=1, meaning it preserves the shape, but not necessarily the distances

x^2 -y^2 = 2 preserves the distances of xy=1