Question

Through what angle should the axes be rotated so that equation $a{x}^2-2\sqrt{3}xy+7y^2=10(x,y)$ may be changed to $3x^2+5y^2=5(x^1,y^1)$

Answer 1

Solution:-

$a{x}^2-2\sqrt{3}xy+7y^2=10(x,y)$

Let the axes rotated by an angle $\theta$.

Let the equation after rotation be-

$a{X}^2-2\sqrt{3}XY+7Y^2=10$

$X=x\cos \theta +y\sin \theta$

$Y=x\sin \theta -y\cos \theta$

$\therefore a{X}^2-2\sqrt{3}XY+7Y^2=10$

$\Rightarrow a{(x\cos \theta +y\sin \theta )}^2-2\sqrt{3}(x\cos \theta +y\sin \theta )(x\sin \theta -y\cos \theta )+7{(x\sin \theta -y\cos \theta )}^2=10$

$\Rightarrow a{x}^2{\cos }^2\theta +a{y}^2{\sin }^2\theta +2axy\sin \theta \cos \theta -2\sqrt{3}{x}^2\cos \theta \sin \theta -2\sqrt{3}xy{\sin }^2\theta +2\sqrt{3}xy{\cos }^2\theta +2\sqrt{3}{y}^2\cos \theta \sin \theta +7x^2{\sin }^2\theta +7y^2{\cos }^2\theta -14xy\sin \theta \cos \theta =10$

$\Rightarrow x^2(a{\cos }^2\theta +7{\sin }^2\theta -2\sqrt{3}\sin \theta \cos \theta )+y^2(a{\sin }^2\theta +7{\cos }^2\theta +2\sqrt{3}\cos \theta \sin \theta )+xy(2a\cos \theta \sin \theta -14\sin \theta \cos \theta -2\sqrt{3}{\sin }^2\theta +2\sqrt{3}{\cos }^2\theta )=10$

$\Rightarrow x^2(a{\cos }^2\theta +7{\sin }^2\theta -2\sqrt{3}\sin \theta \cos \theta )+y^2(a{\sin }^2\theta +7{\cos }^2\theta +2\sqrt{3}\cos \theta \sin \theta )+xy(a\sin 2\theta -7\sin 2\theta +2\sqrt{3}({\cos }^2\theta -{\sin }^2\theta ))=10$

$\Rightarrow x^2(a{\cos }^2\theta +7{\sin }^2\theta -2\sqrt{3}\sin \theta \cos \theta )+y^2(a{\sin }^2\theta +7{\cos }^2\theta +2\sqrt{3}\cos \theta \sin \theta )+xy(a\sin 2\theta -7\sin 2\theta +2\sqrt{3}\cos 2\theta )=10$

As in equation $3x^2+5y^2=5(x^1,y^1)$

Coefficient of $xy$ is 0.

$\therefore a\sin 2\theta -7\sin 2\theta +2\sqrt{3}\cos 2\theta =0$

$\Rightarrow \sin 2\theta (a-7)=-2\sqrt{3}\cos 2\theta$

$\Rightarrow \tan 2\theta =\frac{-2\sqrt{3}}{a-7}$

$\Rightarrow \theta =\frac{{\tan }^{-1}\left(\frac{-2\sqrt{3}}{a-7}\right)}{2}$