Question

The transformed equation of $x^2+2\sqrt{3}xy-y^2-8=0$, when the axes are rotated through an angle $\frac{\pi }{6}$ is

• A. $X^2-Y^2=0$
• B. $X^2-Y^2=4$
• C. $X^2-Y^2=2$
• D. $X^2+Y^2=4$

Answer 1

Answer: (B)

$X^2-Y^2=4$

When axes are rotated through $\frac{\pi }{6}$, equations are

$x=\frac{x\sqrt{3}}{2}-\frac{y}{2}$

$y=\frac{x}{2}+\frac{y\sqrt{3}}{2}$

$x^2+2\sqrt{3xy}-y^2-8=0$

${(\frac{x\sqrt{3}}{2}-\frac{y}{2})}^2+2\sqrt{3}(\frac{x\sqrt{3}}{2}-\frac{y}{2})(\frac{x}{2}+\frac{y\sqrt{3}}{2})-{(\frac{x}{2}+\frac{y\sqrt{3}}{2})}^2-8=0$

$2(x^2-y^2)-8=0$

$x^2-y^2=4$