If $$x\cos \theta - y\sin \theta $$    = $$\sqrt {{x^2} + {y^2}} $$   and $$\frac{{{{\cos }^2}\theta }}{{{a^2}}}$$  + $$\frac{{{{\sin }^2}\theta }}{{{b^2}}}$$  = $$\frac{1}{{{x^2} + {y^2}}}{\text{,}}$$   then the correct relation is?

Solution (By Examveda Team)

$$\eqalign{ & x\cos \theta - y\sin \theta = \sqrt {{x^2} + {y^2}} \,.....(i) \cr & \frac{{{{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}} = \frac{1}{{{x^2} + {y^2}}}\,.....(ii) \cr & \frac{x}{{\sqrt {{x^2} + {y^2}} }}\cos \theta + \frac{{ - y}}{{\sqrt {{x^2} + {y^2}} }}\sin \theta = 1 \cr & {\text{from equation (i)}} \cr & \Rightarrow \sin \theta = \frac{{ - y}}{{\sqrt {{x^2} + {y^2}} }} \cr & \Rightarrow \cos \theta = \frac{x}{{\sqrt {{x^2} + {y^2}} }} \cr & {\text{Put value in equation (ii)}} \cr & \therefore \frac{{{\text{co}}{{\text{s}}^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}} = \frac{1}{{{x^2} + {y^2}}} \cr & \Rightarrow \frac{{{x^2}}}{{\left( {{x^2} + {y^2}} \right){a^2}}} + \frac{{{y^2}}}{{\left( {{x^2} + {y^2}} \right){b^2}}} = \frac{1}{{{x^2} + {y^2}}} \cr & \Rightarrow \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \cr} $$