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  • Two normals at t1 and t2 meet again the parabola y2 = 4ax then the relation between t1 and t2 isOptio

Two normals at t1 and t2 meet again the parabola y2 = 4ax then the relation between t1 and t2 is

Option: 1

t_{1} t_{2}=1


Option: 2

t_{1} -t_{2}=2


Option: 3

t_{1} t_{2}=2


Option: 4

t_{1}+ t_{2}=2


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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Since the normal at t1 meets the parabola at t, so t=-t_{1}-\frac{2}{t_{1}}.

Similarly, t=-t_{1}-\frac{2}{t_{1}}

\begin{array}{l}{\text { Thus, }-t_{1}-\frac{2}{t_{1}}=-t_{2}-\frac{2}{t_{2}}} \\ {\Rightarrow \quad\left(t_{1}-t_{2}\right)=\left(\frac{2}{t_{1}}-\frac{2}{t_{2}}\right)=\frac{2\left(t_{1}-t_{2}\right)}{t_{1} t_{2}}} \\ {\Rightarrow \quad t_{1} t_{2}=2}\end{array}

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