How to find the image of the curve with equation #y=1/x^2#?

enter image source here

1 Answer
Feb 3, 2018

#y=-9/(x^2+4x+4)-1#

Explanation:

All points on the curve #y=1/x^2# will have coordinates of the form:

#(k,1/k^2)#

#[(3,0),(0,-1)][(k),(1/k^2)]=[(3k),(-1/k^2)]+[(-2),(-1)]=[(3k-2),(-1/k^2-1)]#

i.e. #x'=3k-2# , #y'=-1/k^2-1#

Eliminating #k#:

i.e.

#k=(x'+2)/3#

#y'=-1/((x+2)/3)^2-1=-1/(((x^2+4x+4))/9)-1=-9/(x^2+4x+4)-1#

Equation is:

#y=-9/(x^2+4x+4)-1#