A) \[\frac{{{(2x-a)}^{2}}}{{{a}^{2}}}-\frac{2{{y}^{2}}}{{{b}^{2}}}=1\]
B) \[\frac{{{(2x-a)}^{2}}}{{{a}^{2}}}-\frac{4{{y}^{2}}}{{{b}^{2}}}=1\]
C) \[\frac{{{(2x-a)}^{2}}}{{{a}^{2}}}-\frac{8{{y}^{2}}}{{{b}^{2}}}=1\]
D) None of these
Correct Answer: B
Solution :
[b] Let the point P be \[\left( a\text{ }sec\theta ,\text{ }b\,tan\theta \right)\] and A(a, 0), let the midpoint of AP be (h, k) \[\therefore h=\frac{a+a\sec \theta }{2}\,and\,k=\frac{b\tan \theta }{2}\] \[\Rightarrow \frac{2h-a}{a}=\sec \theta \] ???(i) And \[\frac{2k}{b}=\tan \theta \] ???(ii) Squaring and Subtracting (ii) from (i) we get \[\Rightarrow \frac{{{\left( 2h-a \right)}^{2}}}{{{a}^{2}}}-\frac{4{{k}^{2}}}{{{b}^{2}}}=1\] \[\therefore \] Locus of mid points is \[\Rightarrow \frac{{{\left( 2x-a \right)}^{2}}}{{{a}^{2}}}-\frac{4{{y}^{2}}}{{{b}^{2}}}=1\]
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