Pair Of Straight Lines Question 115
Question: The locus of the point $ P(x,y) $ satisfying the relation $ \sqrt{{{(x-3)}^{2}}+{{(y-1)}^{2}}}+\sqrt{{{(x+3)}^{2}}+{{(y-1)}^{2}}}=6 $ is
[Orissa JEE 2002]
Options:
A) Straight line
B) Pair of straight lines
C) Circle
D) Ellipse
Show Answer
Answer:
Correct Answer: B
Solution:
$ \sqrt{{{(x-3)}^{2}}+{{(y-1)}^{2}}}+\sqrt{{{(x+3)}^{2}}+{{(y-1)}^{2}}}=6 $
$ \sqrt{{{(x-3)}^{2}}+{{(y-1)}^{2}}}=6-\sqrt{{{(x+3)}^{2}}+{{(y-1)}^{2}}} $
Squaring both sides.
We get, $ 12x+36=12\sqrt{{{(x+3)}^{2}}+{{(y-1)}^{2}}} $
Again squaring we get the given equation is pair of straight lines.
BETA
