Question

The ends of a rod of length $l$ move on two mutually perpendicular lines. Find the locus of the point on the rod which divides it in the ratio $1:2$.

Answer 1

$letco-ordinatesofPbe(h,k)nowh=\left(\frac{lcos\theta }{1+2}\right)orh=\left(\frac{lcos\theta }{3}\right)\therefore cos\theta =\left(\frac{3h}{l}\right)nowk=\left(\frac{2lsin\theta }{1+2}\right)\therefore sin\theta =\left(\frac{3k}{2l}\right)asweknowsi{n}^2\theta +co{s}^2\theta =1or{\left(\right(\frac{3h}{l}\left)\right)}^2+{\left(\right(\frac{3k}{2l}\left)\right)}^2=1or\left(\frac{9}{l^2}\right)(h^2+(\frac{k^2}{4}\left)\right)=1replacehbyxandkbyywegerlocus{x}^2+\left(\frac{y^2}{4}\right)=\left(\frac{l^2}{9}\right)$