Suppose the axes are to be rotated through an angle $\theta$ so as to remove the $x y$ form from the…
- $(2+\sqrt{3}) x^2+(2-\sqrt{3}) y^2+2 x y=4$
- $(2+\sqrt{3}) x^2+(2+\sqrt{3}) y^2-2 x y=4$
- $x^2+y^2-2(2-\sqrt{3}) x y=4(2-\sqrt{3})$
- $x^2+y^2+2(2+\sqrt{3}) x y=4(2+\sqrt{3})$
Solution
Putting given equation $3 x^2+2 \sqrt{3} x y+y^2=0$ $\begin{aligned} & 3(X \cos \theta-Y \sin \theta)^2+2 \sqrt{3}(X \cos \theta-Y \sin \theta) \\ & (X \sin \theta+Y \cos \theta)+(X \sin \theta+Y \cos \theta)^2=0 \end{aligned}$
$\begin{aligned} \Rightarrow & 3 \cos ^2 \theta X^2+3 \sin ^2 \theta Y^2-6 X Y \sin \theta \cdot \cos \theta \\ & +2 \sqrt{3}\left(X^2 \sin \theta \cos \theta+X Y \cos ^2 \theta-X Y \sin ^2 \theta\right. \\ & \left.-Y^2 \sin \theta \cdot \cos \theta\right)+X^2 \sin ^2 \theta+Y^2 \cos ^2 \theta \\ & +2 X Y \sin \theta \cdot \cos \theta=0\end{aligned}$
$\begin{aligned} & \Rightarrow\left(3 \cos ^2 0+2 \sqrt{3} \sin \theta \cdot \cos \theta+\sin ^2 \theta\right) X^2 \\ &+\left(3 \sin ^2 \theta-2 \sqrt{3} \sin \theta \cdot \cos \theta+\cos ^2 \theta\right) Y^2 \\ &+\left(-6 \sin \theta \cdot \cos \theta+2 \sqrt{3} \cos ^2 \theta-2 \sqrt{3} \sin ^2 \theta\right. \\ &+2 \sin \theta \cdot \cos \theta) X Y=0 \end{aligned}$
To remove $X Y$ term $-3 \sin 2 \theta+2 \sqrt{3} \cos 2 \theta+\sin 2 \theta=0$ $\Rightarrow \tan 2 \theta=\sqrt{3} \Rightarrow \theta=30^{\circ}$ $\therefore$ New transformed equation of $x^2+y^2+2 x y=2$
$\begin{aligned} \begin{aligned}\left(\frac{\sqrt{3} X}{2}-\frac{1}{2} Y\right)^2 & + \\ & \left(\frac{1}{2} X+\frac{\sqrt{3}}{2} Y\right)^2 \\ & +2\left(\frac{\sqrt{3}}{2} X-\frac{1}{2} Y\right)\left(\frac{1}{2} X+\frac{\sqrt{3}}{2} Y\right)=2 \\ \Rightarrow & \frac{3 X^2}{4}+\frac{1}{4} Y^2- \\ & +\frac{\sqrt{3}}{2} X Y+\frac{X^2}{4}+\frac{3 Y^2}{4}+\frac{\sqrt{3}}{2} X Y \\ & +\frac{\sqrt{3}}{2} X^2+X Y-\frac{\sqrt{3}}{2} Y^2=0\end{aligned} \\ \Rightarrow(2+\sqrt{3}) X^2+(2-\sqrt{3}) Y^2+2 X Y=4\end{aligned}$
Asked in: AP EAMCET 2024 (21 May Shift 2)