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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.com/gif.

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

\pi/6,2\pi/3


Option: 2

\pi/6,\pi/3


Option: 3

5\pi/6


Option: 4

\pi/3,2\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+y \cos \theta}, the coefficient of \mathrm{x y} 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 \text {, } }the coefficient of \mathrm{ x y } should be zero, i.e

\mathrm{ -2 \sin \theta \cos \theta+2 \sqrt{3}\left(\cos ^2 \theta-\sin ^2 \theta\right)-2 \sin \theta \cos \theta=0 }

\mathrm{ \Rightarrow \quad \sin 2 \theta-\sqrt{3} \cos 2 \theta=0 }

\mathrm{ \Rightarrow \quad \tan 2 \theta-\sqrt{3} \Rightarrow \theta=\pi/6,2\pi/3. }

Hence option 1 is correct.

 

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