Explanation:
(C) : Given,
The locus of a point and \(\mathrm{p}(\mathrm{x}, \mathrm{y})\) satisfying the equation.
\(\sqrt{(x-2)^{2}+y^{2}}+\sqrt{(x+2)^{2}+y^{2}}=4\)
Now,
\(\sqrt{(x-2)^{2}+y^{2}}=4-\sqrt{(x+2)^{2}+y^{2}}\)
On squaring both side and we get,
\((\mathrm{x}-2)^{2}+\mathrm{y}^{2}=16+(\mathrm{x}+2)^{2}+\mathrm{y}^{2}-2 \times 4 \times \sqrt{(\mathrm{x}+2)^{2}+\mathrm{y}^{2}}\)
\(\Rightarrow \mathrm{x}^{2}+4-4 \mathrm{x}+\mathrm{y}^{2}=16+\mathrm{x}^{2}+4+4 \mathrm{x}+\mathrm{y}^{2}-8 \sqrt{(\mathrm{x}+2)^{2}+\mathrm{y}^{2}}\)
\(\Rightarrow-4 \mathrm{x}=16+4 \mathrm{x}-8 \sqrt{(\mathrm{x}+2)^{2}+\mathrm{y}^{2}}\)
\(\Rightarrow-\mathrm{x}=4+\mathrm{x}-2 \sqrt{(\mathrm{x}+2)^{2}+\mathrm{y}^{2}}\)
\(\Rightarrow-2 \mathrm{x}-4=-2 \sqrt{(\mathrm{x}+2)^{2}+\mathrm{y}^{2}}\)
\(x+2=\sqrt{(x+2)^{2}+y^{2}}\)
Again squaring both side, \(\Rightarrow(\mathrm{x}+2)^{2}=(\mathrm{x}+2)^{2}+\mathrm{y}^{2}\)
\(\Rightarrow \mathrm{y}^{2}=0\)
So, locus of point \(\mathrm{p}(\mathrm{x}, \mathrm{y})\) is a line segment,