A) \[{{t}_{2}}={{t}_{1}}+\frac{2}{{{t}_{1}}}\]
B) \[{{t}_{2}}=-{{t}_{1}}-\frac{2}{{{t}_{1}}}\]
C) \[{{t}_{2}}=-{{t}_{1}}+\frac{2}{{{t}_{1}}}\]
D) \[{{t}_{2}}=-{{t}_{1}}-\frac{2}{{{t}_{1}}}\]
Correct Answer: B
Solution :
Equation of the normal to a parabola \[{{y}^{2}}=4bx\]at point \[\left( bt_{1}^{2},\,2b{{t}_{1}} \right)\] is \[y=-{{t}_{1}}x+2b{{t}_{1}}+bt_{1}^{3}\] As given, it also passes through \[\left( bt_{2}^{2},2b{{t}_{2}} \right)\]then \[2b{{t}_{2}}=-{{t}_{1}}bt_{2}^{2}+2b{{t}_{1}}+bt_{1}^{3}\Rightarrow 2({{t}_{2}}-{{t}_{1}})=-{{t}_{1}}\left( t_{2}^{2}-t_{1}^{2} \right)\]\[\Rightarrow \,\,2{{t}_{2}}-2{{t}_{1}}=-{{t}_{1}}\left( t_{2}^{2}-t_{1}^{2} \right)=-{{t}_{1}}({{t}_{2}}+{{t}_{1}})({{t}_{2}}-{{t}_{1}})\] \[\Rightarrow \,\,2=-{{t}_{1}}({{t}_{2}}+{{t}_{1}})\Rightarrow {{t}_{2}}+{{t}_{1}}=-\frac{2}{{{t}_{1}}}\] \[\Rightarrow \,\,{{t}_{2}}=-{{t}_{1}}-\frac{2}{{{t}_{1}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
