question_answer

Let a circle tangent to the directrix of a parabola \[{{y}^{2}}=2ax\] has its centre coinciding with the focus of the parabola. Then the point of intersection of the parabola and circle is                                          [Orissa JEE 2005]

A)            (a, ?a)                                       

B)            \[(a/2,\ a/2)\]

C)            \[(a/2,\ \pm a)\]                      

D)            \[(\pm a,\ a/2)\]

Correct Answer: C

Solution :

           Given parabola is \[{{y}^{2}}=2ax\]                    \ Focus (a/2, 0) and directrix is given by \[x=-a/2\],                    as circle touches the directrix.                 \ Radius of circle = distance from the point (a/2, 0) to the line                    \[(x=-a/2)\]\[=\frac{\left| \frac{a}{2}+\frac{a}{2} \right|}{\sqrt{1}}=a\]                    \ Equation of circle be \[{{\left( x-\frac{a}{2} \right)}^{2}}+{{y}^{2}}={{a}^{2}}\]             ?..(i)                    also \[{{y}^{2}}=2ax\]                                                  ?..(ii)                    Solving (i) and (ii) we get \[x=\frac{a}{2},\ -\frac{3a}{2}\]                    Putting these values in \[{{y}^{2}}=2ax\] we get                    \[y=\pm a\] and \[x=-3a/2\] gives imaginary values of y.                    \ Required points are\[(a/2,\ \pm a)\].