A) \[x=a\]
B) \[x+a=0\]
C) \[x+2a=0\]
D) \[x+4a=0\]
Correct Answer: B
Solution :
We know that tangent to the parabola at points \[{{t}_{1}}\] and \[{{t}_{2}}\] are \[{{t}_{1}}y=x+at_{1}^{2}\] and \[{{t}_{2}}y=x+at_{2}^{2}.\] Since tangents are perpendicular to the parabola, therefore, \[\frac{1}{{{t}_{1}}}.\frac{1}{{{t}_{2}}}=-1\] or \[{{t}_{1}}{{t}_{2}}=-1\]. We also know that their point of intersection \[=(a{{t}_{1}}{{t}_{2}},\,a({{t}_{1}}+{{t}_{2}}))\] \[=\,(-a,\,a({{t}_{1}}+{{t}_{2}})).\] Thus these points lie on directrix \[x=-\,a\] or \[x+a=0\].
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