If the normal to the parabola y^2 = 4ax inter | vedclass

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(A) The equation of a normal to the parabola $y^2 = 4ax$ is given by $y = mx - 2am - am^3$.If the normal passes through a point on the axis at a dista

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$A.P.$

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(A) The equation of a normal to the parabola $y^2 = 4ax$ is given by $y = mx - 2am - am^3$.If the normal passes through a point on the axis at a distance of $4a$ from the vertex $(0, 0)$,the point is $(4a, 0)$.Substituting $(4a, 0)$ into the equation: $0 = m(4a) - 2am - am^3$.$0 = 2am - am^3$.Since $a 
eq 0$,we have $2m - m^3 = 0$,which implies $m(2 - m^2) = 0$.Thus,the possible slopes are $m = 0$ and $m^2 = 2$,so $m = 0, \sqrt{2}, -\sqrt{2}$.The slopes are $-\sqrt{2}, 0, \sqrt{2}$,which form an $A.P.$ with a common difference of $\sqrt{2}$.

If the normal to the parabola $y^2 = 4ax$ intersects the axis of the parabola at a distance of $4a$ from the vertex,then the slopes of the normal are in which progression?

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