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If the axes be turned through an angle \mathrm{\tan ^{-1} 2}, then the equation \mathrm{4 x y-3 x^2=a^2} becomes

 

Option: 1

\mathrm{x^2-4 y^2=a^2}
 


Option: 2

\mathrm{x^2-4 y^2=2 a^2}
 


Option: 3

\mathrm{x^2+4 y^2=a^2}
 


Option: 4

\mathrm{x^2-2 x y=a^2}


Answers (1)

best_answer

The given equation is \mathrm{4 x y-3 x^2=a^2}

and the angle \mathrm{\alpha=\tan ^{-1} 2 \: or\: \tan \alpha=2}

Hence \mathrm{ \sin \alpha=\frac{2}{\sqrt{5}} \: and \: \cos \alpha=\frac{1}{\sqrt{5}}}
Substituting the appropriate values of \mathrm{x} and \mathrm{y}, we get

\mathrm{ 4\left(x^{\prime} \cos \alpha-y^{\prime} \sin \alpha\right)\left(x^{\prime} \sin \alpha+y^{\prime} \cos \alpha\right)-3\left(x^{\prime} \cos \alpha-y^{\prime} \sin \alpha\right)^2=a^2 }
or
\mathrm{ 4\left[\frac{1}{\sqrt{5}} x^{\prime}-\frac{2}{\sqrt{5}} y^{\prime}\right]\left[\frac{2}{\sqrt{5}} x^{\prime}+\frac{1}{\sqrt{5}} y^{\prime}\right]-3\left[\frac{1}{\sqrt{5}} x^{\prime}-\frac{2}{\sqrt{5}} y^{\prime}\right]^2=a^2 \quad \text { or } \quad x^2-4 y^2=a^2 }

Hence option 1 is correct.

Posted by

Ritika Harsh

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