First slide

Parabola

Question

The locus of the point of intersection of two normals to a parabola which are at right angles to one another is

Moderate

A

$y^{2} = x - 3 a$
B

$y^{2} = a (x - 3 a)$
C

$y^{2} = 2 a (x + 3 a)$
D

$y^{2} = a (x + 3 a)$

Solution

$The equation of the normal to the parabola y^{2} = 4 a x is$ $y = m x - 2 a m - a m^{3}$
$It passes through the point (h, k) if k = m h - 2 a m - a m^{3}$
$or a m^{3} + m (2 a - h) + k = 0 - - - - - (1)$
$Let the roots of the above equation be m_{1}, m_{2}, and m_{3} .$
$Let the perpendicular normals correspond to the values of m_{1}$
$and m_{2} so that m_{1} m_{2} = - 1 .$
$\begin{array}{l} From (1), m_{1} m_{2} m_{3} = - k / a . \\ Since m_{1} m_{2} = - 1, m_{3} = k / a . \\ Since m_{3} is a root of (i), we have \end{array}$
$a {(\frac{k}{a})}^{3} + \frac{k}{a} (2 a - h) + k = 0$
$\begin{array}{l} or k^{2} + a (2 a - h) + a^{2} = 0 \\ or k^{2} = a (h - 3 a) \\ Hence, the locus of (h, k) is \end{array}$
$y^{2} = a (x - 3 a)$

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