Question

The line x cos $\theta$ + y sin $\theta$ = p meets the axes of co-ordinates at A and B respectively. Through A and B lines are drawn parallel to axes so as to meet the perpendicular drawn from origin to given line in P and Q respectively; then show that |PQ| = $\frac{4p|cos2\theta |}{si{n}^{2}2\theta }$

Answer 1

$A(\frac{P}{cos\theta },0),B=\left(\frac{P}{cos\theta }\right)$
Lines through A, B parallel to axes are
$x=\frac{p}{cos\theta },y=\frac{p}{cos\theta }$
These meet the line through origin and .l. to given line i.e., x sin $\theta$ - y cos $\theta$ = 0 in P and Q.
$\therefore P(\frac{P}{cos\theta },\frac{Psin\theta }{co{s}^2\theta }),Q(\frac{Pcos\theta }{si{n}^2\theta },\frac{P}{sin\theta })$
$\therefore P{Q}^2$ = by distance formula
$=P^{2}co{s}^{2}2\theta .[\frac{1}{si{n}^4\theta co{s}^2\theta }+\frac{1}{co{s}^4\theta si{n}^2\theta }]$
$=16p^{2}co{s}^{2}2\theta .\frac{1}{(2sin\theta cos\theta {)}^4}$
$\therefore PQ=\frac{4p|cos2\theta |}{si{n}^{2}2\theta }.$