Pair Of Straight Lines Question 115
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}}}=10 $ is
[Odisha JEE 2002]
Options:
A) Straight line
B) Pair of straight lines
C) Circle.
D) Ellipse is a closed curve with two foci, where the sum of the distances from any point on the curve to the two foci is constant.
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 that the given equation represents a pair of straight lines.
BETA
