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The locus of a point $P(x, y)$ satisfying the equation $\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=4$, is
Options:
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Verified Answer
The correct answer is:
a line segment
Given, equation
$$
\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=4
$$
$\Rightarrow$ Sum of distance of a point $P(x, y)$ w.r.t. points $A(2,0)$ and $B(-2,0)=$ Distance between points $A(2,0)$ and $B((-2,0)$
$$
\Rightarrow \quad A P+B P=A B
$$
$\therefore$ Locus of point $P(x, y)$ is a line segment.
Hence, option (c) is correct.
$$
\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=4
$$
$\Rightarrow$ Sum of distance of a point $P(x, y)$ w.r.t. points $A(2,0)$ and $B(-2,0)=$ Distance between points $A(2,0)$ and $B((-2,0)$
$$
\Rightarrow \quad A P+B P=A B
$$
$\therefore$ Locus of point $P(x, y)$ is a line segment.
Hence, option (c) is correct.
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