find what the following equation becomes when the origin is shifted to the point(1,1)

x²+xy-3y²-y+2=0

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Question

find what the following equation becomes when the origin is
shifted to the point(1,1)
x2+xy-3y2-y+2=0

Manish Bharti · Accepted Answer

When origin is shifted to the point (1,1), then
$x = x' + 1 a n d y = y' + 1$
Substituting the above value in the given equation, we get
$(x' + 1)^{2} + (x' + 1) (y' + 1) + 3 {(y' + 1)}^{2} - (y' + 1) + 2 = 0 x'^{2} + 1 + 2 x' + x' y' + x' + y' + 1 + 3 y'^{2} + 3 + 6 y' - y' - 1 + 2 = 0 x'^{2} + 3 y'^{2} + 3 x' + 6 y' + x' y' + 6 = 0$
Thus, new equation of the curve will be
$x^{2} + 3 y^{2} + 3 x + 6 y + x y + 6 = 0$

Note- When the origin is shifted to the point (h,k) without changing the direction of axes, new equation of the curve can be directly obtained by putting (x+h) in place of x and (y+k) in place of y in the given equation of the curve.

find what the following equation becomes when the origin is shifted to the point(1,1) x2+xy-3y2-y+2=0

find what the following equation becomes when the origin is shifted to the point(1,1)

x²+xy-3y²-y+2=0