Question

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 1

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,

& 2y\dfrac{dy}{dx}=4b \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{4b}{2y} \\\ \end{aligned}$$ Then, $$\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{aligned} & \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{aligned}$$ Dividing by $${{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.