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If the normal at the point \[\left( bt_{1}^{2},2b{{t}_{1}} \right)\] on a parabola, \[{{y}^{2}}=4bx\] meets the curve again at point \[\left( bt_{2}^{2},2b{{t}_{2}} \right)\] then,
(a) \[{{t}_{2}}={{t}_{1}}+\dfrac{2}{{{t}_{1}}}\]
(b) \[{{t}_{2}}=-{{t}_{1}}-\dfrac{2}{{{t}_{1}}}\]
(c) \[{{t}_{2}}=-{{t}_{1}}+\dfrac{2}{{{t}_{1}}}\]
(d) \[{{t}_{2}}={{t}_{1}}-\dfrac{2}{{{t}_{1}}}\]

Answer
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Hint: o solve this question we will first of all determine the equation of normal of given parabola. The equation of normal of parabola of type, \[{{y}^{2}}=4ax\] ar point \[\left( {{x}_{1}},{{y}_{1}} \right)\] is given by,
\[\left( y-{{y}_{1}} \right)=\dfrac{-1}{\dfrac{dy}{dx}}\left( x-{{x}_{1}} \right)\]

Complete step-by-step answer:
Given parabola is, \[{{y}^{2}}=4bx\] this parabola and normal would be of the form.

We have equation of normal of parabola, \[{{y}^{2}}=4ax\] is, \[\left( y-{{y}_{1}} \right)=\dfrac{-1}{\left( \dfrac{dy}{dx} \right)}\left( x-{{x}_{1}} \right)\] at point \[\left( {{x}_{1}},{{y}_{1}} \right)\] - (1)
Given that equation of parabola is, \[{{y}^{2}}=4bx\].
Differentiating both sides with respect to x we get,
\[\begin{align}
  & 2y\dfrac{dy}{dx}=4b \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{4b}{2y} \\
\end{align}\]
Then, \[\dfrac{dy}{dx}=\dfrac{2b}{y}\] - (2)
We are given that the normal is at the point \[\left( bt_{1}^{2},2b{{t}_{1}} \right)\].
Substituting value of \[y=2b{{t}_{1}}\] in equation (2) we get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1\left( 2b \right)}{2b{{t}_{1}}}=\dfrac{1}{{{t}_{1}}}\]
Also the slope of normal is \[\dfrac{-1}{\left( \dfrac{dy}{dx} \right)}\].
\[\Rightarrow \] Slope of normal = \[\dfrac{-1}{\left( \dfrac{1}{{{t}_{1}}} \right)}=-{{t}_{1}}\].
Therefore, equation of normal ar \[\left( bt_{1}^{2},2b{{t}_{1}} \right)\] is,
\[\Rightarrow \left( y-2b{{t}_{1}} \right)=-{{t}_{1}}\left( x-bt_{1}^{2} \right)\] - (3)
Now the point \[\left( bt_{2}^{2},2b{{t}_{2}} \right)\] also lies on the normal. Therefore, point \[\left( bt_{2}^{2},2b{{t}_{2}} \right)\] satisfies (3) we get,
\[\Rightarrow \left( 2b{{t}_{2}}-2b{{t}_{1}} \right)=-{{t}_{1}}\left( bt_{2}^{2}-bt_{1}^{2} \right)\]
Taking 2b common on left we get, and also taking b common on right;
\[\Rightarrow 2b\left( {{t}_{2}}-{{t}_{1}} \right)=-{{t}_{1}}b\left( t_{2}^{2}-t_{1}^{2} \right)\]
Now applying identity \[\left( {{a}_{2}}-{{a}_{1}} \right)\left( {{a}_{2}}+{{a}_{1}} \right)=a_{2}^{2}-a_{1}^{2}\] on the RHS of above equation we get,
\[\Rightarrow 2b\left( {{t}_{2}}-{{t}_{1}} \right)=-{{t}_{1}}b\left( {{t}_{2}}-{{t}_{1}} \right)\left( {{t}_{2}}+{{t}_{1}} \right)\]
Now cancelling \[b\left( {{t}_{2}}-{{t}_{1}} \right)\] on both sides we get,
This can be done as \[b\ne 0\] & \[{{t}_{2}}-{{t}_{1}}\ne 0\].
\[\begin{align}
  & \Rightarrow 2=-{{t}_{1}}\left( {{t}_{2}}+{{t}_{1}} \right) \\
 & \Rightarrow -{{t}_{1}}\left( {{t}_{2}}+{{t}_{1}} \right)=2 \\
 & \Rightarrow -{{t}_{2}}{{t}_{1}}=2+t_{1}^{2} \\
\end{align}\]
Dividing by \[{{t}_{1}}\] we get,
\[\Rightarrow -{{t}_{2}}=\dfrac{2+t_{1}^{2}}{{{t}_{1}}}\]
Multiplying ‘minus’ both sides we get,
\[\Rightarrow {{t}_{2}}=-\dfrac{2}{{{t}_{1}}}-{{t}_{1}}\]
\[\Rightarrow {{t}_{2}}=-{{t}_{1}}-\dfrac{2}{{{t}_{1}}}\], which is option (b).
So, the correct answer is “Option B”.

Note: The possibility of error in this question can be at a point where students directly substitute value of point \[\left( bt_{2}^{2},2b{{t}_{2}} \right)\] in equation of parabola. This would be wrong because this point \[\left( bt_{2}^{2},2b{{t}_{2}} \right)\] is a point of contact normal of parabola. So, we first need to determine the parabola normal of parabola then we can proceed accordingly.