Question

The transformed equation of $9x^2+2\sqrt{3}xy+7y^2=10$ when the axes are rotated through an angle of $\frac{\pi }{6}$ (in the anti clockwise direction) is

• A. $3X^2-Y^2+2\sqrt{2}Y-6=0$
• B. $5X^2+3Y^2=5$
• C. $5X^2-2Y^2=5$
• D. $4X^2+3Y^2=6$

Answer 1

The correct option is B $5X^2+3Y^2=5$

$x=X\cos \frac{\pi }{6}-Y\sin \frac{\pi }{6}=\frac{\sqrt{3}X-Y}{2},y=X\sin \frac{\pi }{6}+Y\cos \frac{\pi }{6}=\frac{X+\sqrt{3}Y}{2}$
Hence the transformed equation is $9{\left(\frac{\sqrt{3}X-Y}{2}\right)}^2+2\sqrt{3}\left(\frac{\sqrt{3}X-Y}{2}\right)\left(\frac{X+\sqrt{3}Y}{2}\right)+7{\left(\frac{X+\sqrt{3}Y}{2}\right)}^2=10$
$\Rightarrow \frac{9}{4}(3X^2-2\sqrt{3}XY+Y^2)+\frac{2\sqrt{3}}{4}(\sqrt{3}{X}^2+2XY-\sqrt{3}{Y}^2)+\frac{7}{4}(X^2+2\sqrt{3}XY+3Y^2)=10\Rightarrow 5X^2+3Y^2=5$