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

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Answer 1 · 2018-02-03T18:04:12

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

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#