Question

By rotating the coordinates axes through ${30}^o$ in anticlockwise sense the equation $x^2+2\sqrt{3}xy-y^2=2a^2$ changes to

• A. $X^2-Y^2=3a^2$
• B. $X^2-Y^2=a$
• C. $X^2-Y^2=2a^2$
• D. $noneofthese$

Answer 1

The correct option is D $noneofthese$

$x=x^{\prime }\cos \theta -y^{\prime }\sin \theta =x^{\prime }\left(\frac{\sqrt{3}}{2}\right)-\frac{y^{\prime }}{2}y=x^{\prime }\sin \theta +y^{\prime }\cos \theta =x^{\prime }\left(\frac{1}{2}\right)+y^{\prime }\left(\frac{\sqrt{3}}{2}\right)x^2+2\sqrt{3}xy-y^2=2a^2\frac{{\left[\sqrt{3}{x}^{\prime }-2y^{\prime }\right]}^2}{4}-\frac{{\left[x^{\prime }-\sqrt{3}{y}^{\prime }\right]}^2}{4}+2\sqrt{3}\frac{\left[\sqrt{3}{x}^{\prime }-y^{\prime }\right]}{2}\frac{{\left[x^{\prime }-\sqrt{3}{y}^{\prime }\right]}^2}{2}=2a^2\frac{2x{}^{\prime 2}-2y{}^{\prime 2}}{4}-\sqrt{3}{x}^{\prime }{y}^{\prime }+\frac{\sqrt{3}}{2}\left[\sqrt{3}{x}^{\prime }-y^{\prime }\right]\left[x^{\prime }+\sqrt{3}y\right]=2a^2-\sqrt{3}{x}^{\prime }{y}^{\prime }+\frac{\sqrt{3}}{2}\left[\sqrt{3}x{}^{\prime 2}-\sqrt{3}y{}^{\prime 2}+2x^{\prime }{y}^{\prime }\right]=2a^{2}2x{}^{\prime 2}-2y{}^{\prime 2}=2a^{2}x{}^{\prime 2}-2y{}^{\prime 2}=2a^2$

$Hence,theoptionDisthecorrectanswer.$