A) \[\]\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}(1-{{e}^{2}})}=1\]
B) \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{a}^{2}}(1-{{e}^{2}})}=1\]
C) \[\frac{{{x}^{2}}}{{{a}^{2}}(1-{{e}^{2}})}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]
D) None of these
Correct Answer: A
Solution :
Let \[A\,(ae,\,\,0)\] and \[B\,(-ae,\,\,0)\] be two given points and \[(h,\,\,k)\] be the coordinates of the moving point P. Now, \[PA+PB=2a\] \[\Rightarrow \,\,\sqrt{{{(h-ae)}^{2}}+{{k}^{2}}}+\sqrt{{{(h+ae)}^{2}}+{{k}^{2}}}=2a\] .....(i) But, we know that \[[{{(h-ae)}^{2}}+{{k}^{2}}]-[{{(h+ae)}^{2}}+{{k}^{2}}]=-4aeh\] .....(ii) Dividing (ii) by (i), we get \[\sqrt{[{{(h-ae)}^{2}}+{{k}^{2}}]}-\sqrt{[{{(h+ae)}^{2}}+{{k}^{2}}]}=-2eh\] .....(iii) Adding (i) and (iii), \[2\sqrt{{{[h-ae)}^{2}}+{{k}^{2}}]}=2\,(a-eh)\] Squaring both sides, we get \[\Rightarrow \,\,{{(h-ae)}^{2}}+{{k}^{2}}={{(a-eh)}^{2}}\,\Rightarrow \,\,\frac{{{h}^{2}}}{{{a}^{2}}}+\frac{{{k}^{2}}}{{{a}^{2}}(1-{{e}^{2}})}=1\] Hence locus of P is \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}(1-{{e}^{2}})}=1\].
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