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  • Find the angle through which the axes must be turned about the origin, so that the equation of the curve <img alt="\mathrm{ x^2+2 \sqrt{3} x y-y^2=2}" src="https://entrancecorner.oncodecogs.co

Find the angle through which the axes must be turned about the origin, so that the equation of the curve \mathrm{ x^2+2 \sqrt{3} x y-y^2=2}  may change in to the form  \mathrm{a x^2+b y^2=1}.

Option: 1

\mathrm{\frac{\pi }{6},\frac{2 \pi}{ 3}}


Option: 2

\mathrm{\frac{\pi}{6}, \frac{\pi}{3}}


Option: 3

\mathrm{\frac{5 \pi}{6}}


Option: 4

\mathrm{\frac{\pi}{3},\frac{2 \pi }{ 3}}


Answers (1)

best_answer

Suppose required angle is \theta. Then on replacing \mathrm{\mathrm{x}~ by ~\mathrm{x} \cos \theta-\mathrm{y} \sin \theta} and \mathrm{y} be\mathrm{x} \sin \theta+ \mathrm{y} \cos \theta, the coefficient of \mathrm{xy} in the new equation should be zero i.e., in the equation

\mathrm{(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=2, }

the coefficient of x y should be zero, i.e.,

\begin{aligned} &\mathrm{ -2 \sin \theta \cos \theta+2 \sqrt{3}\left(\cos ^2 \theta-\sin ^2 \theta\right)-2 \sin \theta \cos \theta=0 }\\ \Rightarrow \quad &\mathrm{ \sin 2 \theta-\sqrt{3} \cos 2 \theta=0} \end{aligned}
\mathrm{\Rightarrow \quad \tan 2 \theta=\sqrt{3} \Rightarrow \quad \theta=\pi / 6,2 \pi / 3 } .
 

Posted by

Suraj Bhandari

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