Question

Transform to parallel axes through the point $(1,-2)$ the equations
(1) $y^2-4x+4y+8=0$,
and (2) $2x^2+y^2-4x+4y=0$.

Answer 1

(1)$y^2-4x+4y+8=\mathrm{0...........}\left(1\right)$

Let the point on the curve with respect to old axis is $(x,y)$

Let the same point on the curve with respect to new axis be $(x^{\prime },y^{\prime })$

The new co-ordinates are shifted by $(1,-2)$

Thus $x=x^{\prime }+1$ and $y=y^{\prime }-2$

Replacing the value of $x$ and $y$ in (1),we get

$(y^{\prime }-2{)}^2-4(x^{\prime }+1)+4(y^{\prime }-2)+8=0$

or, $y^{\prime 2}+4-4y^{\prime }-4x^{\prime }-4+4y^{\prime }-8+8=0$

or, $y^{\prime 2}=4x^{\prime }$

(2)$2x^2+y^2-4x+4y=\mathrm{0..........}\left(2\right)$

Let the point on the curve with respect to old axis is $(x,y)$

Let the same point on the curve with respect to new axis be $(x^{\prime },y^{\prime })$

The new co-ordinates are shifted by $(1,-2)$

Thus $x=x^{\prime }+1$ and $y=y^{\prime }-2$

Replacing the value of $x$ and $y$ in (2),we get

$2(x^{\prime }+1{)}^2+(y^{\prime }-2{)}^2-4(x^{\prime }+1)+4(y^{\prime }-2)=0$

or, $2(x^{\prime 2}+1+2x^{\prime })+(y^{\prime 2}+4-4y^{\prime })-4x^{\prime }-4+4y^{\prime }-8=0$

or, $2x^{\prime 2}+2+4x^{\prime }+y^{\prime 2}-4x^{\prime }-8=0$

or,$2x^{\prime 2}+y^{\prime 2}=6$