Let b, d>0 . locus of all points P(r, θ) for which line P (where, O is origin) cuts line r sin ?

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9.Straight Line

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Let $b, d>0$. The locus of all points $P(r, \theta)$ for which the line $P$ (where, $O$ is the origin) cuts the line $r \sin \theta=b$ in $Q$ such that $P Q=d$ is

A

$(r-d) \sin \theta=b$

B

$(r \pm d) \sin \theta=b$

C

$(r-d) \cos \theta=b$

D

$(r \pm d) \cos \theta=b$

(KVPY-2014)

Solution

(b)

We have, $P \cdot(r, \theta)=(x, y)$

Equation of line $O P$ is

$y=x \tan \theta$

$O P$ cut the line $r \sin \theta=b$

i.e. $y=b$

Given $P Q=d$

$\therefore$ point $P$ is $y=b \pm d \sin \theta$

$\Rightarrow r \sin \theta=b \pm d \sin \theta \Rightarrow(r \pm d) \sin \theta=b$

Hence, locus of $P(r, \theta)$ is $(r \pm d) \sin \theta=b$

Standard 11

Mathematics

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