Vector Algebra Question 61
Question: Let $ x^{2}+3y^{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
Options:
A) $ \pm \hat{j} $
B) $ \pm (\hat{i}+\hat{j}) $
C) $ \pm ,\hat{i} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] Point P lies on $ x^{2}+3y^{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} $ .
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