question_answer

Let \[{{x}^{2}}+3{{y}^{2}}=3\] be the equation of an ellipse in the x-y plane. A and B are two points whose position vectors are \[-\sqrt{3}\hat{i}\] and\[-\sqrt{3}\hat{i}+2\hat{k}\]. Then the position vector of a point P on the ellipse such that \[\angle APB=\pi /4\] is

A) \[\pm \hat{j}\]

B) \[\pm (\hat{i}+\hat{j})\]

C) \[\pm \,\hat{i}\]

D) None of these

Correct Answer: A

Solution :

[a] Point P lies on \[{{x}^{2}}+3{{y}^{2}}=3...(i)\] Now from the diagram, according to the given conditions, \[AP=AB\] or \[{{(x+\sqrt{3})}^{2}}+{{(y-0)}^{2}}=4\] or \[{{(x+\sqrt{3})}^{2}}+{{y}^{2}}=4...(ii)\] Solving (i) and (ii), we get \[x=0\] and \[y=\pm 1\] Hence, point P has position vector \[\pm \hat{j}\].