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प्रश्न
Find what the following equation become when the origin is shifted to the point (1, 1).
x2 + xy − 3y2 − y + 2 = 0
उत्तर
The given equation is x2 + xy − 3y2 − y + 2 = 0.
Substituting \[x = X + 1, y = Y + 1\] in the given equation, we get:
\[\left( X + 1 \right)^2 + \left( X + 1 \right)\left( Y + 1 \right) - 3 \left( Y + 1 \right)^2 - \left( Y + 1 \right) + 2 = 0\]
\[ \Rightarrow X^2 + 1 + 2X + XY + X + Y + 1 - 3 Y^2 - 3 - 6Y - Y - 1 + 2 = 0\]
\[ \Rightarrow X^2 + XY - 3 Y^2 + 3X - 6Y = 0\]
Hence, the transformed equation is \[x^2 + xy - 3 y^2 + 3x - 6y = 0\]
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