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The normal $\mathrm{y=m x-2 a m-a m^3}$ to the parabola $\mathrm{y^2=4 a x }$ subtends a right angle at the vertex if
Option: 1
$\mathrm{m=1}$

Option: 2
$\mathrm{m=\sqrt{2}}$

Option: 3
$\mathrm{m=-\sqrt{2}}$

Option: 4
$\mathrm{m=1/\sqrt{2}}$

Answers (1)

best_answer

Making $\mathrm{y^2=4 a x}$ homogeneous with the help of $\mathrm{y=m x-2 a m-a m^3}$

or $\mathrm{\frac{m x-y}{2 a m+a m^3}=1}$

Then $\mathrm{y^2=4 a x\left(\frac{m x-y}{2 a m+a m^3}\right)}$

$\Rightarrow$ $\mathrm{y^2=\frac{4 x(m x-y)}{2 m+m^3}}$

$\mathrm{\Rightarrow\left(2 m+m^3\right) y^2-4 m x^2+4 x y=0\: \: \: \: \: \: ...(i)}$

$\therefore$ Angle between the lines represented by Eq. (i) is $\mathrm{\pi/2}$

$\therefore$ Coefficient of $\mathrm{x^2+}$ coefficient of $\mathrm{y^2=0}$

$\Rightarrow$ $\mathrm{2 m+m^3-4 m=0}$

$\Rightarrow\: \: \therefore$ $\mathrm{\begin{aligned} m^3-2 m & =0 \\ m^2 & =2, m \neq 0 \end{aligned}}$

$\therefore$ $\mathrm{m= \pm \sqrt{2}}$

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