Question

The point to which the origin should be shifted in order to eliminate $x$ and $y$ terms in the equation $x^2+3y^2-2x+12y+1=0$ is

• A. $(1,-2)$
• B. $(1,3)$
• C. $(-4,3)$
• D. $(-1,2)$

Answer 1

The correct option is A $(1,-2)$

Let origin be shifted to $(h,k)$
$x=x_1+h;y=y_1+k$
$\therefore (x_1+h{)}^2+3(y_1+k{)}^2-2(x_1+h)+12(y_1+k)+1=0$
$2x_{1}h-2x_1=0and6k{y}_1+12y_1=0$
$\Rightarrow h=1andk=-2$
$\therefore (h,k)=(1,-2)$