Question

find orthogonal trajectory of r=a(sec teta+tan teta)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given curve is

$\bf :\longmapsto\:r = a(sec\theta + tan\theta) - - (1)$

On differentiating both sides, we get

$\rm :\longmapsto\:\dfrac{dr}{d\theta} = a(sec\theta \: tan\theta + {sec}^{2}\theta)$

$\rm :\longmapsto\:\dfrac{dr}{d\theta} = a \: sec\theta \: (sec\theta + tan\theta)$

$\rm :\longmapsto\:\dfrac{dr}{d\theta} = \: r \: sec\theta$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: using \: (1)\bigg \}}$

$\rm :\longmapsto\:Replace \: \dfrac{dr}{d\theta} \: by \: - \: {r}^{2}\dfrac{d\theta}{dr}$

$\rm :\longmapsto\:- \: {r}^{2}\dfrac{d\theta}{dr} = r \: sec\theta$

$\rm :\longmapsto\:- \: {r}\dfrac{d\theta}{dr} = \: \dfrac{1}{cos\theta}$

$\rm :\longmapsto\: - \: cos\theta \: d\theta = \dfrac{1}{r} dr$

On integrating both sides, we get

$\displaystyle\rm :\longmapsto\: - \int \: cos\theta \: d\theta = \int \: \dfrac{1}{r} dr$

$\bf :\longmapsto\:c - \: sin\theta = logr$

$\bf :\longmapsto\: \: sin\theta + logr = c$

is the required Orthogonal Trajectory.