A normal chord of the parabola y^2 = 4x subte | vedclass

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(C) For the parabola $y^2 = 4ax$,the equation of a normal at point $t$ is $y + tx = 2at + at^3$. Here $a = 1$,so the normal is $y + tx = 2t + t^3$.If

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$\cot^{-1}(\sqrt{2})$

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(C) For the parabola $y^2 = 4ax$,the equation of a normal at point $t$ is $y + tx = 2at + at^3$. Here $a = 1$,so the normal is $y + tx = 2t + t^3$.If a chord joining $t_1$ and $t_2$ subtends a right angle at the vertex $(0,0)$,then the product of the slopes of the lines joining the vertex to the points is $-1$. The slope of the line joining $(0,0)$ and $(at_1^2, 2at_1)$ is $m_1 = \frac{2}{t_1}$ and $m_2 = \frac{2}{t_2}$.Thus,$\frac{2}{t_1} 	imes \frac{2}{t_2} = -1 \Rightarrow t_1t_2 = -4$.Since the chord is a normal,$t_2 = -t_1 - \frac{2}{t_1}$.Substituting $t_2 = -\frac{4}{t_1}$,we get $-\frac{4}{t_1} = -t_1 - \frac{2}{t_1}$ $\Rightarrow t_1^2 = 2$ $\Rightarrow t_1 = \sqrt{2}$ (for acute angle).The slope of the normal is $m = -t_1 = -\sqrt{2}$.The angle $	heta$ with the $x$-axis is given by $	an(\pi - 	heta) = |m| = \sqrt{2}$.Therefore,$	an(	heta) = \frac{1}{\sqrt{2}}$,which implies $	heta = 	an^{-1}(\frac{1}{\sqrt{2}}) = \cot^{-1}(\sqrt{2})$.

$A$ normal chord of the parabola $y^2 = 4x$ subtending a right angle at the vertex makes an acute angle $\theta$ with the $x$-axis. Then $\theta$ is equal to:

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