If $${p_1},\,{p_2}$$  are the lengths of the normals drawn from the origin on the lines $$x\,\cos \,\theta + y\,\sin \,\theta = 2a\,\cos \,4\theta $$       and $$x\,\sec \,\theta + y\,{\text{cosec }}\theta = 4a\,\cos \,2\theta $$       and respectively, and $$mp_1^2 + np_2^2 = 4{a^2}.$$     Then :

Solution :
$$\eqalign{ & p_1^2 = 4{a^2}{\cos ^2}4\theta \cr & p_2^2 = \frac{{16{a^2}{{\cos }^2}2\theta }}{{{{\sec }^2}\theta + {\text{cose}}{{\text{c}}^2}\theta }} \cr & = 16{a^2}{\cos ^2}2\theta \,{\cos ^2}\theta \,{\sin ^2}\theta \cr & = {a^2}{\sin ^2}4\theta \cr & \therefore \,p_1^2 + 4p_2^2 = 4{a^2} \cr} $$