Is (x-h)^4(y-k)=Some Constant Also an Equation of a Rectangular Hyperbola?

[gliteringstar]

the general equation for rectangular hyperbola with vertical and horizontal asymptotes is given as :

(x-h)(y-k)= some constant

Is the following also an equation of rectangular hyperbola

(x-h)^4(y-k)=some constant ?

I am trying to find the shape of this curve,is it similar to that of rectangular hyperbola?

 

[gliteringstar]

...to be more precise...

the cartesian equation of rectangular hyperbola in question is given by:

xy=a^2

If we have it as (x-h)^4(y-k)=a^2,will the shape of graph of this rectangular hyperbola change?...the centre is different of course...it has shifted to (h,k) with the two parts of the graph lying in the newly formed first and third quadrant...

someone please reply!

 

[HallsofIvy]

With the fourth power, it will NOT be a hyperbola at all- though it may look somewhat like one. A "hyperbola", is defined as being one of the "conic sections" and so its equation must be second degree, not fifth degree as you give.

 

[gliteringstar]

thank you :)

But how do we find the graph of such a function?and if not the graph,can we get some idea about point of intersection of graph of this function with any of the axes?

 

[CellCoree]

Yes, the equation (x-h)^4(y-k)=some constant is also an equation of a rectangular hyperbola. The shape of this curve will still be similar to that of a rectangular hyperbola, with the only difference being that the curve will be more stretched or flattened depending on the value of the exponent on (x-h). The general equation (x-h)(y-k)= some constant represents a rectangular hyperbola with its center at (h,k), while the equation (x-h)^4(y-k)=some constant represents a rectangular hyperbola with its center at (h,k) and a higher degree of stretching or flattening. Both equations will have the same vertical and horizontal asymptotes, and the same general shape of a rectangular hyperbola.