question_answer

The locus of a point whose difference of distance from points (3, 0) and (-3,0) is 4, is     [MP PET 2002]

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 \[P\,(h,\,\,k).\] Given that \[PA-PB=4\] \[\sqrt{{{(h-3)}^{2}}+{{k}^{2}}}-\sqrt{{{(h+3)}^{2}}+{{k}^{2}}}=4\] \[\Rightarrow \,\,\sqrt{{{(h-3)}^{2}}+{{k}^{2}}}=4+\sqrt{{{(h+3)}^{2}}+{{k}^{2}}}\] Squaring both sides, we get \[{{(h-3)}^{2}}+{{k}^{2}}=16+{{(h+3)}^{2}}+{{k}^{2}}+8\sqrt{{{(h+3)}^{2}}+{{k}^{2}}}\] \[\Rightarrow \,\,\,{{h}^{2}}+9-6h+{{k}^{2}}=16+{{h}^{2}}+9+6h+{{k}^{2}}\]\[+\,8\sqrt{{{(h+3)}^{2}}+{{k}^{2}}}\] \[\Rightarrow \,\,-6h=16+6h+8\sqrt{{{(h+3)}^{2}}+{{k}^{2}}}\] \[\Rightarrow \,\,-8\,\sqrt{{{(h+3)}^{2}}+{{k}^{2}}}=12h+16\] Again, squaring both sides, we get \[64\,({{h}^{2}}+9+6h+{{k}^{2}})=144{{h}^{2}}+256+2.16.12h\] \[\Rightarrow \,\,4\,({{h}^{2}}+9+6h+{{k}^{2}})=9{{h}^{2}}+16+24h\] \[\Rightarrow \,\,4{{h}^{2}}+36+24h+4{{k}^{2}}=9{{h}^{2}}+16+24h\] \[\Rightarrow \,\,5{{h}^{2}}-4{{k}^{2}}=20\,\,\Rightarrow \,\,\frac{{{h}^{2}}}{4}-\frac{{{k}^{2}}}{5}=1\] Hence, the locus of point P is \[\frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{5}=1\].