If the normals drawn from any point to the parabola cut the line $x = 2 a$ in points whose ordinates are in arithmetical progression, prove that the tangents of the angles which the normals make with the axis are in geometrical progression.

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Solution

The equation of normal in parametric form is

$y = - t x + 2 a t + a t^{3}$

$a t^{3} + (2 a - x) t - y = 0$

This is cubic in $t$

${\Rightarrow t}_{1} + t_{2} + t_{3} = 0 t_{1} + t_{3} = - t_{2} . . . . . . . . . . . (i)$

At $x = 2 a$

$y = a t^{3}$

So the ordinates at $x = 2 a$ are $a t_{1}^{3}, a t_{2}^{3}, a t_{3}^{3}$

Given $2 a t_{2}^{3} = a t_{1}^{3} + a t_{3}^{3}$

$2 t_{2}^{3} = t_{1}^{3} + t_{3}^{3} 2 t_{2}^{3} = (t_{1} + t_{3}) (t_{1}^{2} + t_{3}^{2} - t_{1} t_{3})$

substituting $(i)$

$2 t_{2}^{3} = - t_{2} (t_{1}^{2} + t_{3}^{2} - t_{1} t_{3}) - 2 t_{2}^{2} = t_{1}^{2} + t_{3}^{2} - t_{1} t_{3} - 2 t_{2}^{2} = {(t_{1} + t_{3})}^{2} - 3 t_{1} t_{3} - 2 t_{2}^{2} = {(t_{2})}^{2} - 3 t_{1} t_{3} - 3 t_{2}^{2} = - 3 t_{1} t_{3} t_{2}^{2} = t_{1} t_{3}$

Slope of normal $= tan ϕ = t$

${(- tan ϕ_{2})}^{2} = (- tan ϕ_{1}) (- tan ϕ_{3}) {tan}^{2} ϕ_{2} = tan ϕ_{1} tan ϕ_{3}$

Clearly $tan ϕ_{1}, tan ϕ_{2}, tan ϕ_{3}$ are in $G . P$

Hence proved.

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