$\begin{array}{1 1}(a)\;m_1+m_2+m_3=0\\(b)\;m_1m_2+m_2m_3+m_3m_1=\large\frac{2a-h}{a}\\(c)\;m_1m_2m_3=\large\frac{-k}{a}\\(d)\;\text{All of these}\end{array}$

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P(h,k) for a parabola $y^2=4ax$ is a co-normal point.

We know that for a parabola $y^2=4ax$ normal is given by

$y=mx-2am-am^3$

(h,k) will satisfy the equation.

$k=mh-2am-am^3$

$am^3+m(2a-h)+k=0$

This is cubic in m,so it has three roots $m_1,m_2,m_3$

$\therefore m_1+m_2+m_3=0$-----(1)

$m_1m_2+m_2m_3+m_3m_1=\large\frac{2a-h}{a}$-------(2)

$m_1m_2m_3=\large\frac{-k}{a}$------(3)

equation(1),(2)&(3) are the three conditions.

Hence (d) is the correct answer.

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