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 :
[b] Equation of the normal to a parabola \[{{y}^{2}}=4bx\] at point \[\left( b{{t}^{2}}_{1},2b{{t}_{1}} \right)\]is \[y=-{{t}_{1}}x+2b{{t}_{1}}+b{{t}^{3}}_{1}\] As given, it also passes through \[\left( b{{t}^{2}}_{2},2b{{t}_{2}} \right)\] then \[2b{{t}_{2}}=-{{t}_{1}}b{{t}^{2}}_{2}+2b{{t}_{1}}+b{{t}^{3}}_{1}\] \[\Rightarrow 2=-{{t}_{1}}({{t}_{2}}+{{t}_{1}})\Rightarrow {{t}_{2}}+{{t}_{1}}=-\frac{2}{{{t}_{1}}}\] \[\Rightarrow {{t}_{2}}=-{{t}_{1}}-\frac{2}{{{t}_{1}}}\]
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