is there any relation between y1x graph and hyperbola if so

is there any relation between y=1/x graph and hyperbola? if so than explain how?

Solution

y = 1/x is a hyperbola. You probably learned that a hyperbola has the standard form of:

x^2/a^2 - y^2/b^2 = 1. (So it\'s second degree equation in 2 variables).

The reason that y=1/x doesn\'t look like that is because it has been rotated 45 degrees from the standard position. In fact, it is really the hyperbola:
x^2/ 2 - y^2/2 = 1. This hyperbola is centered at the origin and the foci are on the x-axis. It is symmetric with respect to the x-axis (so it \"opens\" to the left and right).
If you take this hyperbola and rotate it 45 degrees counterclockwise, you can show that the equation becomes
xy = 1. Which can be written y = 1/x. (The \"cross term\" xy indicates it\'s been rotated).

The hyperbola belongs to the same class with the circle, ellipse, and parabola. They are known as \"conic sections\" (because they can be generated by intersecting a cone and a plane) and all have second degree equations. The most general equation of this class is:
ax^2 + bxy + cy^2 + dx +ey + f = 0. The standard hyperbola, for example, would result if b=d=e=0, f = -1, and c is negative. The parabola y = x^2 would result if a=-1, e=1, and b=c=d=f=0, and so on .

Given a function f(x), the inverse of this (if one exists) is a function g such that
g( f(x) ) = x. So to talk about an inverse you need to be dealing with some function already. For example, the inverse of ln x is e^x.
If f(x) = 1/x, then the inverse function of f happens to be itself ! Namely
g(x) = 1/x. You see, for example, that f(2)=1/2 and then g(1/2) = 2. In general, g( f(x) ) would = g( 1/x ) = 1/(1/x) = x.