A line L has intercepts a and b on the coordinate axes. When the coordinate axes are rotated through an angle$\alpha$keeping the origin fixed, the same line L has intercepts p and q on the new axes. Then

$\text{ Suppose the axes are rotated in the anti-clockwise direction through an angle }\alpha$

$\text{ . The equation of the line }\mathrm{L}\text{ with respect to the old axes is given by }$

$x/a+y/b=1$

$\text{ To find the equation of }L\text{ with respect to the new axes, we }$

$\text{ replace }\mathrm{x}\text{ by }\mathrm{xcos}\alpha -\mathrm{ysin}\alpha \text{ and }\mathrm{y}\text{ by }\mathrm{xsin}\alpha +\mathrm{ycos}\alpha \text{, so that the equation of }\mathrm{L}\text{ with respect to the new axes is }$

$\frac{1}{a}(x\cos \alpha -y\sin \alpha )+\frac{1}{b}(x\sin \alpha +y\cos \alpha )=1$

$\text{ Since }p,q\text{ are the intercepts made by this line on the coordinate axes, we have }$

$\text{ on putting }(p,o)\text{ and then }(o,q)$

$1/p=(1/a)\cos \alpha +(1/b)\sin \alpha \text{ and }1/q=-(1/a)\sin \alpha +(1/b)\cos \alpha$

Elliminate $\alpha$ will get,

$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$