A) 0
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{6}\]
Correct Answer: B
Solution :
Any line through .origin is y = mx. Since, it is a tangent to \[{{y}^{2}}=4a(x-a),\]it will cut it in two coincident points. So, roots of \[{{m}^{2}}{{x}^{2}}-4ax+4{{a}^{2}}\]are equal. \[\therefore \] Product of slope \[=-1\,i/e.,{{b}^{2}}-4ac=0\] \[\Rightarrow \]\[16{{a}^{2}}-16{{a}^{2}}{{m}^{2}}=0\] \[\Rightarrow \]\[{{m}^{2}}=1\]or \[m=1,-1\] Hence, required angle is right angle i.e.,\[\frac{\pi }{2}.\]
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