A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{2}\]
Correct Answer: C
Solution :
We know that tangent to \[{{y}^{2}}=4ax\]is \[y=mx+\frac{a}{m}.\] \[\Rightarrow \] tangent to \[{{y}^{2}}=4x\]is \[y=mx+\frac{1}{m}\] Since, tangent passes through (1, 4) \[\therefore \] \[4=m+\frac{1}{m}\] \[\Rightarrow \] \[{{m}^{2}}-4m+1=0\] (whose roots are\[{{m}_{1}},{{m}_{2}}\]) \[\therefore \] \[{{m}_{1}}+{{m}_{2}}=4\] and \[{{m}_{1}}{{m}_{2}}=1\] and \[|{{m}_{1}}-{{m}_{2}}|\,=\sqrt{{{({{m}_{1}}+{{m}_{2}})}^{2}}-4{{m}_{1}}{{m}_{2}}}\] \[=\sqrt{12}=2\sqrt{3}\] Thus, the angle between tangent \[\tan \theta =\left| \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{2\sqrt{3}}{1+1} \right|=\sqrt{3}\] \[\Rightarrow \] \[\theta =\frac{\pi }{3}\]
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