Consider f(x) =A normal to y = f(x) at x = also passes through the point

Engineering

Mathematics

Tangent and Normal

Question

Consider f(x) = $\tan^{- 1} (\sqrt{\frac{1 + \sin (x)}{1 - \sin (x)}}) x \in (0, \frac{π}{2})$

A normal to y = f(x) at x = $\frac{π}{6}$ also passes through the point

$(\frac{π}{4}, 0)$

$(0, \frac{2 π}{3})$

(0, 0)

$(\frac{π}{6}, 0)$

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$f (x) = \tan^{- 1} (\frac{\sin \frac{x}{2} + \cos \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}) = \tan^{- 1} (\tan (\frac{π}{4} + \frac{x}{2})) = \frac{π}{4} + \frac{x}{2}$

$f' (x) = \frac{1}{2}$

$f (\frac{π}{6}) = \tan^{- 1} \sqrt{3} = \frac{π}{3}$

Equation of normal at $(\frac{π}{6}, \frac{π}{3})$ is

$y - \frac{π}{3} = - 2 (x - \frac{π}{6}) \Rightarrow 2 x + y - \frac{2 π}{3} = 0$

which passes through $(0, \frac{2 π}{3}) .$