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  • If the normal at P to the parabola y2 = 4ax cuts again at Q then the distance between P and Q is?Option: 1 <img alt="a|t_2+t_1|\sqrt{

If the normal at P to the parabola y2 = 4ax cuts again at Q then the distance between P and Q is?

Option: 1

a|t_2+t_1|\sqrt{(t_2-t_1)^2+4 }


Option: 2

a|t_2+t_1|\sqrt{(t_2+t_1)^2+4 }


Option: 3

a|t_2-t_1|\sqrt{(t_2-t_1)^2+4 }


Option: 4

a|t_2-t_1|\sqrt{(t_2+t_1)^2+4 }


Answers (1)

best_answer

 

 

Normal at t1 meets the parabola again at t2 -

Normal at t1 meets the parabola again at t2

\\ {\text { Equation of Normal at } \mathrm{P} \equiv\left(\mathrm{at}_{1}^{2}, 2 \mathrm{at}_{1}\right) \text { to the parabola } \mathrm{y}^{2}=4 \mathrm{ax\;\;is}}\\ \\ {\mathrm{y}=-\mathrm{t}_{1 \mathrm{X}}+2 \mathrm{at}_{1}+\mathrm{at}_{1}^{3}} \\\\ {\text { It meets the parabola again at } \mathrm{Q} \equiv\left(\mathrm{at}_{2}^{2}, 2 \mathrm{at}_{2}\right)} \\\\ {\therefore 2 \mathrm{at}_{2}=-\mathrm{at}_{1} \mathrm{t}_{2}^{2}+2 \mathrm{at}_{1}+\mathrm{at}_{1}^{3}} \\\\ {\Rightarrow 2 \mathrm{a}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)\left[2+\mathrm{at}_{1}\left(\mathrm{t}_{2}^{2}-\mathrm{t}_{1}^{2}\right)=0\right.} \\ {\Rightarrow \mathrm{a}\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)\left[2+\mathrm{t}_{1}\left(\mathrm{t}_{2}+\mathrm{t}_{1}\right)\right]=0} \\ {\therefore a\left(t_{2}-t_{1}\right)=0} \\ {\therefore 2+\mathrm{t}_{1}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)=0} \\\\ \mathbf{{\mathrm{t}_{2}=-\mathrm{t}_{1}-\frac{2}{\mathrm{t}_{1}}}}

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\begin{aligned} \text { Thus, } P Q &=\sqrt{\left(at^2_2-at^2_1\right)^{2}+\left(2at_2-2at_1\right)^{2}} \\ &=a|t_2-t_1|\sqrt{(t_2+t_1)^2+4 } \end{aligned}

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Pankaj

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