| The locus of a point whose difference of distance from points (3, 0) and \[(-\,\,3,0)\]is 4, is |
A) \[\frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{5}=1\]
B) \[\frac{{{x}^{2}}}{5}-\frac{{{y}^{2}}}{4}=1\]
C) \[\frac{{{x}^{2}}}{2}-\frac{{{y}^{2}}}{3}=1\]
D) \[\frac{{{x}^{2}}}{3}-\frac{{{y}^{2}}}{2}=1\]
Correct Answer: A
Solution :
| Let the point be \[(x,y).\] |
| Then. \[\sqrt{{{(x-3)}^{2}}+{{y}^{2}}}-\sqrt{{{(x+3)}^{2}}+{{y}^{2}}}=4\] |
| \[\Rightarrow \]\[\sqrt{{{(x-3)}^{2}}+{{y}^{2}}}=4+\sqrt{{{(x+3)}^{2}}+{{y}^{2}}}\] |
| On squaring both sides, we get |
| \[{{(x-3)}^{2}}+{{y}^{2}}=16+{{(x+3)}^{2}}+{{y}^{2}}+8\sqrt{{{(x+3)}^{2}}+{{y}^{2}}}\]\[\Rightarrow \]\[-\,\,6x-6x-16=8\sqrt{{{(x+3)}^{2}}+{{y}^{2}}}\] |
| \[\Rightarrow \] \[-\,\,12x-16=8\sqrt{{{(x+3)}^{2}}+{{y}^{2}}}\] |
| \[\Rightarrow \] \[-\,\,3x-4=2\sqrt{{{(x+3)}^{2}}+{{y}^{2}}}\] |
| Again, on squaring both sides, we get |
| \[9{{x}^{2}}+16+24x=4\,[{{(x+3)}^{2}}+{{y}^{2}}]\] |
| \[\Rightarrow \] \[9{{x}^{2}}+16+24x=4\,[{{x}^{2}}+9+6x+{{y}^{2}}]\] |
| \[\Rightarrow \] \[9{{x}^{2}}+16+24x=4{{x}^{2}}+36+24x+4{{y}^{2}}\] |
| \[\Rightarrow \] \[5{{x}^{2}}-4{{y}^{2}}=20\]\[\Rightarrow \] \[\frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{5}=1\] |
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