If the origin is shifted to remove the first degree terms from the equation $2 x^2-3 y^2+4 x y+4 x+4 y-14=0$…
If the origin is shifted to remove the first degree terms from the equation $2 x^2-3 y^2+4 x y+4 x+4 y-14=0$ then, with respect to this new co-ordinate system, the transformed equation of $x^2+y^2-3 x y+4 y+3=0$ is
$x^2+y^2-3 x y-2 x+y+6=0$
$x^2+y^2-3 x y-2 x+7 y+3=0$
$x^2+y^2-3 x y-2 x+y+4=0$
$x^2+y^2-3 x y-2 x+7 y+4=0$
Solution
$2 x^2-3 y^2+4 x y+4 x+y-14=0$
Let $x=\mathrm{X}+h, y=\mathrm{Y}+k$
Then $2(\mathrm{X}+\mathrm{h})^2-3(\mathrm{Y}+k)^2+4(\mathrm{X}+\mathrm{h})(\mathrm{Y}+k)+4(\mathrm{X}+\mathrm{h})$ $+4(\mathrm{Y}+k)-14=0$
$2 \mathrm{X}^2-3 \mathrm{Y}^2+4 \mathrm{XY}+(4+4 h+4 k) \mathrm{X}+(4-6 k+4 h) \mathrm{Y}$
$+2 h^2-3 k^2+4 h k+4 h+4 k-14=0$
To remove first degree terms
$\begin{aligned}
& 4+4 h+4 k=0 \text { and } 4-6 k+4 h=0 \\
& \Rightarrow k=0, h=-1 \\
& \therefore x=\mathrm{X}-1, y=\mathrm{Y}
\end{aligned}$
Now, transformation of $x^2+y^2-3 x y+4 y+3=0$ is
$\begin{aligned}
& (\mathrm{X}-1)^2+\mathrm{Y}^2-3(\mathrm{X}-1) \mathrm{Y}+4 \mathrm{Y}+3=0 \\
& \Rightarrow \mathrm{X}^2+\mathrm{Y}^2-3 \mathrm{XY}-2 \mathrm{X}+7 \mathrm{Y}+4=0
\end{aligned}$