Question

If the axes are shifted to $(-2,-3)$ and then rotated through $\frac{\pi }{4}$ in anticlockwise direction, then transformed equation of $x^2-y^2+2x+4y=0$ is

• A. $8x+12y-2\sqrt{2}xy-21\sqrt{2}=0$
• B. $8x-12y-2\sqrt{2}xy-21\sqrt{2}=0$
• C. $8x+12y+2\sqrt{2}xy-21\sqrt{2}=0$
• D. $8x+12y+2\sqrt{2}xy+21\sqrt{2}=0$

Answer 1

The correct option is A $8x+12y-2\sqrt{2}xy-21\sqrt{2}=0$

As $(0,0)$ is shifted to $(-2,-3)$
$x=X-2,y=Y-3$
Where $(x,y)$ are the original coordinates and $(X,Y)$ are transformed coordinates.
$\because$ After origin shifting axes are rotated through $\frac{\pi }{4}$
$X=X^{\prime }\cos \frac{\pi }{4}-Y^{\prime }\sin \frac{\pi }{4}=\frac{X^{\prime }-Y^{\prime }}{\sqrt{2}}X=X^{\prime }\cos \frac{\pi }{4}+Y^{\prime }\sin \frac{\pi }{4}=\frac{X^{\prime }+Y^{\prime }}{\sqrt{2}}$
Where $(X^{\prime },Y^{\prime })$ are the rotated coordinates,
Therefore,
$x=X-2=\frac{X^{\prime }-Y^{\prime }}{\sqrt{2}}-2y=Y-3=\frac{X^{\prime }+Y^{\prime }}{\sqrt{2}}-3$
Now, finding the transformed equation for
$x^2-y^2+2x+4y=0$
Putting the respective values of $(x,y)$, we get
${(\frac{X^{\prime }-Y^{\prime }}{\sqrt{2}}-2)}^2-{(\frac{X^{\prime }+Y^{\prime }}{\sqrt{2}}-3)}^2+2(\frac{X^{\prime }-Y^{\prime }}{\sqrt{2}}-2)+4(\frac{X^{\prime }+Y^{\prime }}{\sqrt{2}}-3)=0\Rightarrow -\frac{2}{\sqrt{2}}(X^{\prime }-Y^{\prime })+\frac{10}{\sqrt{2}}(X^{\prime }+Y^{\prime })-2X^{\prime }{Y}^{\prime }-21=0\Rightarrow 8X^{\prime }+12Y^{\prime }-2\sqrt{2}{X}^{\prime }{Y}^{\prime }-21\sqrt{2}=0$
So, the transformed equation is
$8x+12y-2\sqrt{2}xy-21\sqrt{2}=0$