Question

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

Answer 1

Let, $\theta$ be the angle through which the axes are turned anti-clockwise, then

$\tan \theta =2$

$\Rightarrow \sin \theta =\frac{2}{\sqrt{5}}$ and $\cos \theta =\frac{1}{\sqrt{5}}$.

Let, $P(X,Y)$ be the new co-ordinate of any point $p(x,y)$, when the axes are turned through an angle $\theta$.

Then, $x=X\cos \theta -Y\sin \theta$ and $y=X\sin \theta +Y\cos \theta$

or, $x=\frac{X-2Y}{\sqrt{5}}$ and $y=\frac{2X+Y}{\sqrt{5}}$.

Putting these values of $x,y$ in the given equation $4xy-3x^2=a^2$, we have

$4\left(\frac{X-2Y}{\sqrt{5}}\right)\left(\frac{2X+Y}{\sqrt{5}}\right)-3{\left(\frac{X-2Y}{\sqrt{5}}\right)}^2$ $=a^2$

or, $4(X-2Y)(2X+Y)-3(X-2Y{)}^2=5a^2$

or, $4(2X^2-3XY-2Y^2)-3(X^2-4XY+4Y^2)=5a^2$

or, $5X^2-20Y^2=5a^2$

or, $X^2-4Y^2=a^2$.