The locus of the point P(x, y) satisfying the relation {(x-3)²+(y-1)² +{(x+3)²+(y-1)²=6 is

Correct Answer is : pair of straight lines

Correct Answer is : pair of straight lines

Test Index

Manipal Entrance Test 2019 Math Paper

Question : 44 of 70

Marks: +1, -0

P(x, y)

\sqrt{(x-3)^2+(y-1)^2}

+\sqrt{(x+3)^2+(y-1)^2}=6

Solution:

Given that,

\sqrt{(x-3)^2+(y-1)^2}

+\sqrt{(x+3)^2+(y-1)^2}=6

\Rightarrow\sqrt{(x-3)^2+(y-1)^2}

=6-\sqrt{(x+3)^2+(y-1)^2}

On squaring both sides, we get

(x-3)^2+(y-1)^2

=36+(x+3)^2+(y-1)^2

-12\sqrt{(x+3)^2+(y-1)^2}

\Rightarrow x^2-6x+9+(y^2-2y+1)

=36+x^2+6x+9+y^2-2y+1

-12\sqrt{(x+3)^2+(y-1)^2}

\Rightarrow(x+3)=\sqrt{(x+3)^2+(y-1)^2}

On squaring both sides of above equation

\Rightarrow(x+3)^2=[(x+3)^2+(y-1)^2]

\Rightarrow(y-1)^2=0

⇒ Which represents a pair of straight lines.

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