If normals are drawn from a point $P (h, k)$ to the parabola $y^{2} = 4 a x$ , then the sum of the intercepts which the normals cut off from the axis of the parabola is

(h + a)

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2 (h - a)

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2 (h + a)

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none of these

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Solution

The correct option is C $2 (h + a)$
Equation of normal $y = m x - 2 a m - a m^{3}$
Let $m_{1}, m_{2}, m_{3}$ be the slopes of normals
Put $y = 0$ , we get
$x_{1} = 2 a + a m_{1}^{2}$
$x_{2} = 2 a + a m_{2}^{2}$
$x_{3} = 2 a + a m_{3}^{2}$
where $x_{1}, x_{2}, x_{3}$ are the intercepts on the axis of the parabola.
$\Rightarrow x_{1} + x_{2} + x_{3} = 6 a + a (m_{1}^{2} + m_{2}^{2} + m_{3}^{2})$
The normal passes through (h, k).
$\Rightarrow a m^{3} + (2 a - h) m + k = 0$
which has roots $m_{1}, m_{2}, m_{3}$ which are slopes of the normals.
$\Rightarrow m_{1} + m_{2} + m_{3} = 0$
and $m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = \frac{2 a - h}{a}$
$\Rightarrow m_{1}^{2} + m_{2}^{2} + m_{3}^{2} = (m_{1} + m_{2} + m_{3})^{2} - 2 (m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1})$
$= - \frac{2 (2 a - h)}{a}$
$\Rightarrow x_{1} + x_{2} + x_{3} = 6 a - 2 (2 a - h) = 2 (h + a)$

Suggest Corrections

Similar questions

P (h, k)

y^{2} = 4 a x

2 (h + a)

A

y^{2} = 4 a x

x^{2} = 4 b y

y^{2} = 4 a x

x^{2} = 4 b y

y^{2} = 4 a x

x^{2} = 4 b y

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