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\[(-1,0)\]You need to login to perform this action.
You will be redirected in
3 sec