Question

The point where the origin has to be shifted so that the equation $y^2+4y+8x-2=0$ will not contain $y$ and the constant term is

• A. $(\frac{3}{4},-2)$
• B. $(\frac{3}{5},-2)$
• C. $(2,4)$
• D. $(1,-4)$

Answer 1

The correct option is A $(\frac{3}{4},-2)$

Let the origin be shifted to $(h,k)$.Then, $x=X+h$ and $y=Y+k$. Substituting $x=X+h$ and $y=Y+k$ in the equation $y^2+4y+8x-2=0$, we get
$(Y+k{)}^2+4(Y+k)+8(X+h)-2=0$

$\Rightarrow Y^2+2kY+k^2+4Y+4k+8X+8h-2=0$

$\Rightarrow Y^2+(4+2k)Y+8X+(k^2+4k+8h-2)=0$
For this equation to be free from the term containing $Y$ and the constant term, we must have

$4+2k=0$ and $k^2+4k+8h-2=0$
$k=-2$ and $h=\frac{3}{4}$
Hence, the origin has to be shifted at the point $(\frac{3}{4},-2)$.