Question

If the axes be turned through an angle ${\tan }^{-1}2$, what does the equation $4xy-3x^2=a^2$ become?

Answer 1

According to new axis,

$x=x^{\prime }\cos \theta -y^{\prime }\sin \theta$

$y=x^{\prime }\sin \theta +y^{\prime }\cos \theta$

Here $\theta ={\tan }^{-1}2$

So, $\cos \theta =\cos \left({\tan }^{-1}2\right)=\cos \left({\cos }^{-1}\right(\frac{1}{\sqrt{5}}\left)\right)=\frac{1}{\sqrt{5}}$

and, $\sin \theta =\sin \left({\tan }^{-1}2\right)=\sin \left({\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)=\frac{2}{\sqrt{5}}$

So, $x$ becomes $\frac{x^{\prime }}{\sqrt{5}}-\frac{2y^{\prime }}{\sqrt{5}}$ and $y$ becomes $\frac{2x^{\prime }}{\sqrt{5}}+\frac{y^{\prime }}{\sqrt{5}}$

So, the equation $4xy-3x^2=a^2$ becomes,

$4(\frac{x^{\prime }}{\sqrt{5}}-\frac{2y^{\prime }}{\sqrt{5}})$$(\frac{2x^{\prime }}{\sqrt{5}}+\frac{y^{\prime }}{\sqrt{5}})$$-3{(\frac{x^{\prime }}{\sqrt{5}}-\frac{2y^{\prime }}{\sqrt{5}})}^2=a^2$

or, $4(\frac{2x^{\prime 2}}{5}-\frac{3x^{\prime }{y}^{\prime }}{5}-\frac{2y^{\prime }}{5})-3(\frac{x^{\prime 2}}{5}+\frac{4y^{\prime 2}}{5}-\frac{4x^{\prime }{y}^{\prime }}{5})=a^2$

or, $x^{\prime 2}-4y^{\prime 2}=a^2$