Question

Find the new position of origin so that equation $x^2+4 x+8 y-2=0$ will not contain a term in $x$ and the costant term.

• A. $\left(\frac{3}{4}, 4\right)$
• B. $\left(\frac{3}{4},-2\right)$
• C. $\left(-2, \frac{3}{4}\right)$
• D. $\left(-2,-\frac{3}{4}\right)$

Answer 1

Answer: (C)

$x=x+h \& y=y+k$
$(x+h)^2+4(x+h)+8(y+k)-2=0$
$x^2+(2 h+4) x+8 y+\left(h^2+4 h+8 k-2\right)=0$
$\therefore 2 h+4=0 \& h^2+4 h+8 k-2=0$
$h=-2 \,\,\,\,4-8+8 k-2=0$
$k=\frac{6}{8}=\frac{3}{4}$
$\left(-2, \frac{3}{4}\right)$