question_answer

The point of intersection of tangents at the ends of the latus-rectum of the parabola \[{{y}^{2}}=4x\] is equal to          [Pb. CET 2003]

A)            (1, 0)                                         

B)            (?1, 0)

C)            (0, 1)                                         

D)            (0, ?1)

Correct Answer: B

Solution :

           Equation of the tangent at \[({{x}_{1}},\,{{y}_{1}})\] on the parabola \[{{y}^{2}}=4ax\] is \[y{{y}_{1}}=2a(x+{{x}_{1}})\]                    \[\therefore \] In this case, \[a=1\]                   The co-ordinates at the ends of the latus rectum of the parabola \[{{y}^{2}}=4x\] are \[L(1,\,2)\] and \[{{L}_{1}}(1,\,-2)\]                    Equation of tangent at L and \[{{L}_{1}}\] are \[2y=2(x+1)\] and\[-2y=2(x+1)\], which gives \[x=-1\],\[y=0\]. Thus, the required point of intersection is (?1, 0).