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).You need to login to perform this action.
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