A) \[\frac{1}{3}\]
B) \[\frac{1}{\sqrt{3}}\]
C) \[\sqrt{3}\]
D) \[3\]
Correct Answer: D
Solution :
The locus of the point of intersection of tangents to the parabola \[{{y}^{2}}=4ax\] inclined at an angle \[\alpha \] to each other is\[{{\tan }^{2}}\alpha {{(x+a)}^{2}}={{y}^{2}}-4ax\] Given equation of Parabola \[{{y}^{2}}=4x\{a=1\}\] Point of intersection (-2, -1) \[{{\tan }^{2}}\alpha {{(-2+1)}^{2}}={{(-1)}^{2}}-4\times 1\times (-2)\] \[\Rightarrow \]\[{{\tan }^{2}}\alpha =9\]\[\Rightarrow \]\[\tan \alpha =\pm 3\]\[\Rightarrow \]\[|\tan \alpha |=3\]
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