Question

Find the locus of a point $O$ when the three normal drawn from it on to the parabola $y^{2} = 4 a x$ are such that two of them make equal angles with the given line $y = m x + c$ .

Accepted Answer

Given parabola $y^{2} = 4 a x \dots (i)$

The equation of normal to it will be

$y = m x - 2 a m - a m^{3}$ (in slope form) which passes through $O (h, k)$

$\Rightarrow k = m h - 2 a m - a m^{3} \Rightarrow a m^{3} + (2 a - h) m + k = 0$

Let $m_{1}, m_{2}, m_{3}$ be the roots of the above equation

$\Rightarrow m_{1} + m_{2} + m_{3} = 0$ , $m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = \frac{2 a - h}{a}, m_{1} m_{2} m_{3} = \frac{- k}{a} .$

given line $y = m x + c . . . . . (i i)$

Let $θ$ be the angle made by (ii) with the positive direction of x-axis Let $θ_{1}, θ_{2}$ be the angles made by two normal with the positive direction of the $x$ -axis and these normals make equal angles with line (ii) clearly from the figure.

$θ_{2} - θ = θ - θ_{1} \Rightarrow 2 θ = θ_{1} + θ_{2}$

$\tan 2 θ = \tan (θ_{1} + θ_{2}) \Rightarrow \frac{2 \tan θ}{1 - \tan^{2} θ} = \frac{\tan θ_{1} + \tan θ_{2}}{1 - \tan θ_{1} \tan θ_{2}}$

$\frac{2 \tan θ}{1 - \tan^{2} θ} = \frac{\tan θ_{1} + \tan θ_{2}}{1 - \tan θ_{1} + \tan θ_{2}} = λ \in R$

$\Rightarrow \frac{2 m}{1 - m^{2}} = \frac{m_{1} + m_{2}}{1 - m_{1} m_{2}} = λ$

$\Rightarrow \frac{- m_{3}}{1 - (\frac{- k}{a m_{3}})} = λ \Rightarrow a m_{3}^{2} + a m_{3} λ + k λ = 0 . . . . . (i i i)$

$m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = \frac{2 a - h}{a}$

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

$\Rightarrow - a m_{3}^{3} - k = (2 a - h) m_{3}$

$\Rightarrow a m_{3}^{3} + (2 a - h) m_{3} + k = 0 . . . . . (i v)$

multiply (iii) with $m_{3}$

$\Rightarrow a m_{3}^{3} + a λ m_{3}^{2} + k λ m_{3} = 0 . . . . . . (v)$

$(v) - (i v) \Rightarrow a λ m_{3}^{2} + (k λ - 2 a + h) m_{3} - k = 0 . . . . . . . (v i)$

from (iii) and (vi) By cross-ratio method

$\begin{matrix} k λ - 2 a + h & - k & a λ & (k λ - 2 a + h) \\ a λ & k λ & a & a λ \end{matrix}$

$\frac{m^{2}}{k λ (k λ - 2 a + h) + a λ k} = \frac{m}{- a k - a k λ^{2}} = \frac{1}{a^{2} λ^{2} - a k λ + 2 a^{2} - h a}$

$\frac{m^{2}}{k^{2} λ^{2} - a k λ + h k λ} = \frac{m}{- a k (1 + λ^{2})} = \frac{1}{a (a λ^{2} - k λ + 2 a - h)}$

$\frac{k (k λ^{2} - a λ + h λ)}{a (a λ^{2} - k λ + 2 a - h)} = {(\frac{- a k (1 + λ^{2})}{a (a λ^{2} - k λ + 2 a - h)})}^{2}$

$\frac{k^{2} {(1 + λ^{2})}^{2}}{{(a λ^{2} - k λ + 2 a - h)}^{2}} = \frac{k (k λ^{2} - a λ + h λ)}{a (a λ^{2} - k λ + 2 a - h)}$

$k^{2} a {(1 + λ^{2})}^{2} = k (k λ^{2} - a λ + h λ) (a λ^{2} - k λ + 2 a - h)$

Locus of $(h, k)$ is

$a y {(1 + λ^{2})}^{2} = (y λ^{2} - a λ + x λ) (a λ^{2} - λ y + 2 a - x) \dots (v i i)$

Coefficient of $x^{2} = - λ = a$ ,Coefficient of $y^{2} = - λ^{3} = b$ and Coefficient of $x y = - 2 λ^{2} \Rightarrow h = - λ^{2} \Rightarrow h^{2} = λ^{4} = a b$

for convenience

Take $a = 1, λ = 1$ (since $(a, λ \in R a n d λ \neq 0)$

$\Rightarrow 4 y = (y - 1 + x) (1 - y + 2 - x)$

$4 y = (x + y - 1) (3 - (x + y))$

$4 y = 3 (x + y) - 3 - {(x + y)}^{2} + x + y$

$4 y = 4 x + 4 y - 3 - x^{2} - y^{2} - 2 x y$

$Δ = |\begin{matrix} 1 & 1 & - 2 \\ 1 & 1 & 0 \\ - 2 & 0 & 3 \end{matrix}| = 1 (3) - 1 (3) - 2 (+ 2) = - 4 \neq 0$

$\Rightarrow$ Equation $(v i i)$ represents parabola

Locus of $O$ is a parabola

S L Loney Solutions for Chapter: The Parabola (Continued), Exercise 2: EXAMPLES XXX

S L Loney Mathematics Solutions for Exercise - S L Loney Solutions for Chapter: The Parabola (Continued), Exercise 2: EXAMPLES XXX

Questions from S L Loney Solutions for Chapter: The Parabola (Continued), Exercise 2: EXAMPLES XXX with Hints & Solutions

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