Find the new equation of the curve (x-2)^2=y(y-1)^2(x−2)2=y(y−1)2 by transforming to parallel axes through the point (2, 1)(2,1)?

1 Answer
Shwetank Mauria
Feb 13, 2018

New equation is (x-4)^2=(y-1)(y-2)^2(x−4)2=(y−1)(y−2)2

Explanation:

Transforming to parallel axes through (x_1,y_1)(x1,y1) means the transformation x->x+x_1x→x+x1 and y->y+y_1y→y+y1

Hence here it is x->x+2x→x+2 and y->y+1y→y+1

and equation (x-2)^2=y(y-1)^2(x−2)2=y(y−1)2

becomes (x+2-2)^2=(y+1)(y+1-1)^2(x+2−2)2=(y+1)(y+1−1)2

or x^2=y^2(y+1)x2=y2(y+1)

The graph of given equation (x-2)^2=y(y-1)^2(x−2)2=y(y−1)2 appears as

graph{((x-2)^2-y(y-1)^2)(x-2)(y-1)=0 [-10, 10, -5, 5]}

and graph of given equation x^2=y^2(y+1)x2=y2(y+1) appears as

graph{x^2=y^2(y+1) [-10, 10, -5, 5]}

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