A) \[\pm {{\tan }^{-1}}\left( \frac{4}{29} \right)\]
B) \[\pm {{\tan }^{-1}}\left( \frac{5}{49} \right)\]
C) \[\pm {{\tan }^{-1}}\left( \frac{10}{49} \right)\]
D) \[\pm {{\tan }^{-1}}\left( \frac{8}{29} \right)\]
Correct Answer: C
Solution :
Given that, \[x={{t}^{2}}+1\] and \[y={{t}^{2}}-t-6\] \[\therefore \] \[\frac{dx}{dt}=2t\] and \[\frac{dy}{dt}=2t-1\] Now, \[\frac{dy}{dt}=\frac{dy/dt}{dx/dt}=\frac{2t-1}{2t}\] when, it meet x-axis, then \[{{t}^{2}}-t-6=0\] \[\Rightarrow \] \[(t+2)\,(t-3)=0\] \[\Rightarrow \] \[t=3,\,-2\] \[\therefore \] \[{{\left( \frac{dy}{dx} \right)}_{at\,t=3}}=\frac{6-1}{6}=\frac{5}{6}={{m}_{1}}(\text{say})\] and \[{{\left( \frac{dy}{dx} \right)}_{\text{at}\,t=-2}}=\frac{5}{4}={{m}_{2}}(\text{say})\] Required angle\[=\pm {{\tan }^{-1}}\left\{ \left| \frac{\frac{5}{6}-\frac{5}{4}}{1+\frac{25}{24}} \right| \right\}\] \[=\pm {{\tan }^{-1}}\left\{ \frac{10}{49} \right\}\]
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