The sum of the intercepts which the normals cut off from the axis is $2 (h + a)$ .

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Solution

Equation of normal in slope form is
$y = m x - 2 a m - a m^{3}$
It passes through $(h, k)$
$= m h - 2 a m - a m^{3} a m^{3} + (2 a - h)) m + k = 0$
This is cubic in $m$
$∴ m_{1} + m_{2} + m_{3} = - \frac{0}{a} = 0 m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = \frac{2 a - h}{a} m_{1} m_{2} m_{3} = - \frac{k}{a}$
$y = m x - 2 a m - a m^{3}$
$y = 0 \Rightarrow x = 2 a + a m^{2}$
So the sum of intercept from the axis
$S = 2 a + a m_{1}^{2} + 2 a + a m_{2}^{2} + 2 a + a m_{3}^{2}$
$S = 6 a + a (m_{1}^{2} + m_{2}^{2} + m_{3}^{2}) S = 6 a + a {{(m_{1} + m_{2} + m_{3})}^{2} - 2 (m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1})} S = 6 a + a {{(0)}^{2} - 2 (\frac{2 a - h}{a})} S = 6 a - 4 a + 2 h S = 2 (a + h)$
Hence proved.

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