Question

The point to which the origin should be shifted in order to remove the $x$ and $y$ terms in the equation $14x^2-4xy+11y^2-36x+48y+41=0$ is

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

Answer 1

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

Let the origin be shifted to the point $(h,k)$.

Let $(x,y)$ and $(x^{\prime },y^{\prime })$ are the coordinates of a point in the old and new system respectively, then $x=x^{\prime }+h,y=y^{\prime }+k$.

So, the transformed equation is

$14{(x^{\prime }+h)}^2-4(x^{\prime }+h)(y^{\prime }+k)+11{(y^{\prime }+k)}^2-36(x^{\prime }+h)+48(y^{\prime }+k)+41=0\Rightarrow 14{x^{\prime }}^2+28x^{\prime }h+14h^2-4x^{\prime }{y}^{\prime }-4x^{\prime }k-4y^{\prime }h-4hk+11{y^{\prime }}^2+11k^2+22y^{\prime }k-36x^{\prime }-36h+48y^{\prime }+48k+41=0\Rightarrow 14{x^{\prime }}^2+11{y^{\prime }}^2+x^{\prime }(28h-4k-36)+y^{\prime }(-4h+22k+48)+(14h^2-4hk-36h+48k+41)-4x^{\prime }{y}^{\prime }=0$

In order to remove the first degree term, we must have

$28h-4k-36=0$ and $-4h+22k+48=0$

$\Rightarrow h=1$ and $k=-2$

$\therefore$ the origin must be shifted to the point $(1,-2)$ to remove the $1^{st}$ degree terms from the given equation.