Find the transformed equation of the curve $x^2+2 \sqrt{3} x y-y^2=8$, when the axes are rotated through an…
Find the transformed equation of the curve $x^2+2 \sqrt{3} x y-y^2=8$, when the axes are rotated through an angle $\frac{\pi}{3}$.
$x^2+y^2+2 \sqrt{3} x y=8$
$x^2+y^2-2 \sqrt{3} x y=8$
$x^2-y^2+2 \sqrt{3} x y=8$
$x^2-y^2-2 \sqrt{3} x y=8$
Solution
Given equation,
$x^2+2 \sqrt{3} x y-y^2=8$
Since, the axes are rotated through an angle $\pi / 3$.
$\therefore(x, y)$ replaced by
$\left(x \cos \frac{\pi}{3}-y \sin \frac{\pi}{3}, x \sin \frac{\pi}{3}+y \cos \frac{\pi}{3}\right)$
i.e. $\left(\frac{x-\sqrt{3} y}{2}, \frac{\sqrt{3} x+y}{2}\right)$ in the given equation.
$\therefore\left(\frac{x-\sqrt{3} y}{2}\right)^2+2 \sqrt{3}\left(\frac{x-\sqrt{3} y}{2}\right)$
$\left(\frac{\sqrt{3} x+y}{2}\right)-\left(\frac{\sqrt{3} x+y}{2}\right)^2=8$
$\begin{aligned} \Rightarrow \frac{1}{4}\left(x^2-2 \sqrt{3} x y\right. & +3 y^2+6 x^2-4 \sqrt{3} x y \\ & \left.-6 y^2-3 x^2-2 \sqrt{3} x-y^2\right)=8\end{aligned}$
$\Rightarrow \quad \frac{1}{4}\left(4 x^2-8 \sqrt{3} x y-4 y^2\right)=8$
$\Rightarrow \quad x^2-2 \sqrt{3} x y-y^2=8$