question_answer

    The point of intersection of tangents at the ends of the latus-rectum of the parabola \[{{x}^{2}}+{{y}^{2}}-18x+16y+120=0\] is equal to :

A)  (1,0)                                     

B) \[{{x}^{2}}+4x+2y=0\]

C) \[2y+3=0\]                                         

D) \[3y=2\]

Correct Answer: B

Solution :

                Equation to 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\[2y=(2x+1)\]and\[-2y=2(x+1)\]which gives \[x=-1,\text{ }y=0\] thus the required point of intersection is\[(-1,0)\]