Let b, d 0. The locus of all points P(r, th | vedclass

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(B) Let the coordinates of point $P$ be $(r, 	heta)$. In Cartesian coordinates,$P = (r \cos 	heta, r \sin 	heta)$.The line $OP$ passes through the

Vedclass · Accepted Answer

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

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(B) Let the coordinates of point $P$ be $(r, 	heta)$. In Cartesian coordinates,$P = (r \cos 	heta, r \sin 	heta)$.The line $OP$ passes through the origin and has the equation $y = x 	an 	heta$.The line $r \sin 	heta = b$ is equivalent to $y = b$ in Cartesian coordinates.Let $Q$ be the point of intersection of line $OP$ and the line $y = b$.Since $Q$ lies on $y = b$,its $y$-coordinate is $b$. Since $Q$ lies on $OP$,its polar distance from the origin $OQ$ satisfies $OQ \sin 	heta = b$,so $OQ = \frac{b}{\sin 	heta}$.Given $PQ = d$,the distance $OP = OQ \pm d = \frac{b}{\sin 	heta} \pm d$.Thus,$r = \frac{b}{\sin 	heta} \pm d$.Multiplying by $\sin 	heta$,we get $r \sin 	heta = b \pm d \sin 	heta$,which simplifies to $(r \mp d) \sin 	heta = b$.Since $d$ is a constant,the locus is $(r \pm d) \sin 	heta = b$.

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

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