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In the $x y$-plane, three distinct lines $l_{1}, l_{2}, l_{3}$ concur at a point $(\lambda, 0)$. Further the lines $l_{1}, l_{2}, l_{3}$ are normals to the parabola $y^{2}=6 x$ at the points $A=\left(x_{1}, y_{1}\right), B=\left(x_{2}, y_{2}\right), C=\left(x_{3}, y_{3}\right)$ respectively. Then we have -
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Verified Answer
The correct answer is:
$\lambda>3$
Any normal
$\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^{3}$ Here $\mathrm{a}=3 / 2$
through $(\lambda, 0)$
$\begin{array}{l}
0=\mathrm{m} \lambda-2 \mathrm{am}-\mathrm{am}^{3} \\
\mathrm{~m}=0, \quad \lambda=2 \mathrm{a}+\mathrm{am}^{2} \\
\mathrm{~m}^{2}=\frac{\lambda}{\mathrm{a}}-2>0 \\
\lambda>2 \mathrm{a} \Rightarrow \lambda>3
\end{array}$
$\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^{3}$ Here $\mathrm{a}=3 / 2$
through $(\lambda, 0)$
$\begin{array}{l}
0=\mathrm{m} \lambda-2 \mathrm{am}-\mathrm{am}^{3} \\
\mathrm{~m}=0, \quad \lambda=2 \mathrm{a}+\mathrm{am}^{2} \\
\mathrm{~m}^{2}=\frac{\lambda}{\mathrm{a}}-2>0 \\
\lambda>2 \mathrm{a} \Rightarrow \lambda>3
\end{array}$
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