The point to which the origin is to be shifted to remove the first degree terms from the equation $2 x^2+4 x…
The point to which the origin is to be shifted to remove the first degree terms from the equation $2 x^2+4 x y-6 y^2+2 x+8 y+1=0$ is
$\left(\frac{7}{8}, \frac{3}{8}\right)$
$\left(\frac{-7}{8}, \frac{-3}{8}\right)$
$\left(\frac{-7}{8}, \frac{3}{8}\right)$
$\left(\frac{7}{8}, \frac{-3}{8}\right)$
Solution
$2 x^2+4 x y-6 y^2+2 x+8 y+1=0$
is in the form of
$
\begin{aligned}
& a x^2+2 h x y+b y^2+2 g x+2 f y+c=0, \text { we get } \\
& a=2, h=2, b=-6, g=1, f=4, c=1
\end{aligned}
$
Required point
$
\begin{aligned}
& =\left(\frac{b g-f h}{h^2-a b}, \frac{a f-g h}{h^2-a b}\right)=\left(\frac{-6-8}{4+12}, \frac{8-2}{4+12}\right) \\
& =\left(\frac{-14}{16}, \frac{6}{16}\right)=\left(\frac{-7}{8}, \frac{3}{8}\right)
\end{aligned}
$