A) \[x=-a\]
B) \[x=-\frac{a}{2}\]
C) \[x=0\]
D) \[x=\frac{a}{2}\]
Correct Answer: C
Solution :
\[\alpha =\frac{a{{t}^{2}}+a}{2},\,\,\beta =\frac{2at+0}{2}\Rightarrow \,2\alpha =a{{t}^{2}}+a,\,at=\beta \] \ \[2\alpha =a.\,\frac{{{\beta }^{2}}}{{{a}^{2}}}+a\] or \[2a\alpha ={{\beta }^{2}}+{{a}^{2}}\] \ The locus is \[{{y}^{2}}=\frac{4a}{2}\left( x-\frac{a}{2} \right)\] \[=4b(x-b),\,\left( b=\frac{a}{2} \right)\] \ Directrix is \[(x-b)\,+b=0\] or \[x=0\].You need to login to perform this action.
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