Three normals drawn from any point to the par | vedclass

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(B) The equation of the normal in parametric form is $y = -tx + 2at + at^3$.Rearranging,we get $at^3 + (2a - x)t - y = 0$.This is a cubic equation in

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$G.P.$

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(B) The equation of the normal in parametric form is $y = -tx + 2at + at^3$.Rearranging,we get $at^3 + (2a - x)t - y = 0$.This is a cubic equation in $t$,so the sum of the roots $t_1 + t_2 + t_3 = 0$,which implies $t_1 + t_3 = -t_2$ $(i)$.At $x = 2a$,the equation becomes $at^3 + (2a - 2a)t - y = 0$,so $y = at^3$.The ordinates at $x = 2a$ are $at_1^3, at_2^3, at_3^3$.Given these are in $A$.$P$.,we have $2at_2^3 = at_1^3 + at_3^3$,or $2t_2^3 = t_1^3 + t_3^3$.Using the identity $t_1^3 + t_3^3 = (t_1 + t_3)(t_1^2 + t_3^2 - t_1t_3)$,we substitute $t_1 + t_3 = -t_2$:$2t_2^3 = -t_2(t_1^2 + t_3^2 - t_1t_3)$.Dividing by $-t_2$ (assuming $t_2 
eq 0$),we get $-2t_2^2 = t_1^2 + t_3^2 - t_1t_3$.Since $t_1^2 + t_3^2 = (t_1 + t_3)^2 - 2t_1t_3 = (-t_2)^2 - 2t_1t_3 = t_2^2 - 2t_1t_3$,we have:$-2t_2^2 = t_2^2 - 2t_1t_3 - t_1t_3 = t_2^2 - 3t_1t_3$.$-3t_2^2 = -3t_1t_3$,so $t_2^2 = t_1t_3$.The slope of the normal is $m = -t$,so $	an \phi = -t$.Thus,$(-	an \phi_2)^2 = (-	an \phi_1)(-	an \phi_3)$,which simplifies to $	an^2 \phi_2 = 	an \phi_1 	an \phi_3$.Therefore,$	an \phi_1, 	an \phi_2, 	an \phi_3$ are in $G.P.$

Three normals drawn from any point to the parabola $y^2 = 4ax$ cut the line $x = 2a$ in points whose ordinates are in arithmetical progression. Then the tangents of the angles which the normals make with the axis of the parabola are in:

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