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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3 (h + a)

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2 (h + a)

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4 (h + a)

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none of these

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Solution

The correct option is D $2 (h + a)$
Equation of normal is $y = m x - 2 a m - a m^{3}$

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 intercepts on the axis of the parabola.

The normal passes through $(h, k)$

$∴ a m^{3} + (2 a - h) m + k = 0$

Now, $m_{1} + m_{2} + m_{3} = 0$ and $m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = \frac{2 a - h}{a}$

$\Rightarrow {m_{1}}^{2} + {m_{2}}^{2} + {m_{3}}^{2} = {(m_{1} + m_{2} + m_{3})}^{2} - 2 (m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1}) = - 2 (\frac{2 a - h}{a})$

$x_{1} + x_{2} + x_{3} = 2 a + a {m_{1}}^{2} + 2 a + a {m_{2}}^{2} + 2 a + a {m_{3}}^{2} = 6 a + 2 a [m_{1}^{2} + m_{2}^{2} + m_{3}^{2}]$

$\Rightarrow x_{1} + x_{2} + x_{3} = 6 a - 2 (2 a - h) = 2 (h + a)$

Suggest Corrections

Similar questions

P (h, k)

y^{2} = 4 a x

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

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