Coordinate Geometry Question 46
Question: Two points A and B have coordinates (1, 0) and (-1, 0) respectively and Q is a point which satisfies the relation $ AQ-BQ= $
$ \pm 1. $ The locus of Q is
[MP PET 1986]
Options:
A) $ 12x^{2}+4y^{2}=3 $
B) $ 12x^{2}-4y^{2}=3 $
C) $ 12x^{2}-4y^{2}+3=0 $
D) $ 12x^{2}+4y^{2}+3=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
According to the given condition $ \sqrt{{{(x-1)}^{2}}+y^{2}}-\sqrt{{{(x+1)}^{2}}+y^{2}}=\pm 1 $ On squaring both sides, we get $ 2x^{2}+2y^{2}+1=2\sqrt{{{(x-1)}^{2}}+y^{2}}.\sqrt{{{(x+1)}^{2}}+y^{2}} $ Again on squaring, we get $ 12x^{2}-4y^{2}=3 $ .
BETA
