In the XY-plane,three distinct lines l_1, l_2 | vedclass

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(B) The equation of the parabola is $y^2 = 6x$,so $4a = 6$,which gives $a = \frac{3}{2}$.The equation of a normal to the parabola $y^2 = 4ax$ at a poi

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$\lambda > 3$

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(B) The equation of the parabola is $y^2 = 6x$,so $4a = 6$,which gives $a = \frac{3}{2}$.The equation of a normal to the parabola $y^2 = 4ax$ at a point with slope $m$ is $y = mx - 2am - am^3$.Substituting $a = \frac{3}{2}$,the equation of the normal becomes $y = mx - 3m - \frac{3}{2}m^3$.Since the normal passes through the point $(\lambda, 0)$,we substitute these coordinates into the equation:$0 = m\lambda - 3m - \frac{3}{2}m^3$.For $m 
eq 0$ (as the normal is not the axis of the parabola),we can divide by $m$:$0 = \lambda - 3 - \frac{3}{2}m^2$.Rearranging for $m^2$,we get $\frac{3}{2}m^2 = \lambda - 3$,or $m^2 = \frac{2}{3}(\lambda - 3)$.For three distinct normals to exist,there must be three distinct real values for $m$. Since $m^2 = \frac{2}{3}(\lambda - 3)$,for $m$ to have three distinct real roots (including $m=0$ for the axis),we require $\lambda - 3 > 0$,which implies $\lambda > 3$.

In the $XY$-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 $y^2=6x$ at the points $A=(x_1, y_1)$,$B=(x_2, y_2)$,and $C=(x_3, y_3)$ respectively. Then,we have:

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