A) \[{{30}^{o}}\]
B) \[{{45}^{o}}\]
C) \[{{60}^{o}}\]
D) \[{{90}^{o}}\]
Correct Answer: D
Solution :
We know that tangent to \[{{y}^{2}}=4ax\] is \[y=mx+\frac{a}{m}\] \[\therefore \] Tangent to \[{{y}^{2}}=4x\] is \[y=mx+\frac{1}{m}\]. Since, tangent passes through \[(1-,-6)\]. \[\therefore \] \[-6=-m+\frac{1}{m}\] \[\Rightarrow \] \[{{m}^{2}}-6m-1=0\] Whose roots are \[{{m}_{1}}\] and \[{{m}_{2}}\]. \[\therefore \] \[{{m}_{1}}+{{m}_{2}}=6\] and \[{{m}_{1}}\,{{m}_{2}}=-1\] Now, \[|{{m}_{1}}-{{m}_{2}}|=\sqrt{{{({{m}_{1}}+{{m}_{2}})}^{2}}-4{{m}_{1}}{{m}_{2}}}\] \[=\sqrt{36+4}\] \[=2\sqrt{10}\] Thus, angle between tangent is \[\tan \theta =\left| \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right|\] \[=\left| \frac{2\sqrt{10}}{1-1} \right|=\infty \] \[\Rightarrow \] \[\theta ={{90}^{o}}\]
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