Question

If p and q are the lengths of perpendicular from the origin to the lines $xcos\theta -ysin\theta =kcos2\theta$ and $xsec\theta +ycosec\theta =k$, respectively.

Prove that $p^2+4q^2=k^2$

Answer 1

p =Length of perpendicular from origin (0, 0) to the line $xcos\theta -ysin\theta -kcos2\theta =0$

Then, $p=\frac{|0.cos\theta +0.(-sin\theta )-kcos2\theta |}{\sqrt{co{s}^2\theta +si{n}^2\theta }}$

[ $\because$ the perpendicular distance from $(x_1,y_1)$ to the line ax + by + c = 0 is$\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}$]

$\therefore$ p = k cos $2\theta$ . . . (i)

and q = Length of perpendicular from origin (0, 0) to the line $xsec\theta +ycosec\theta -k=0$

Then, $q=\frac{|0.sec\theta +0.cosec\theta -k|}{\sqrt{se{c}^2\theta +cose{c}^2\theta }}$

$=\frac{x}{\sqrt{\frac{1}{co{s}^2\theta }+\frac{1}{si{n}^2\theta }}}=\frac{k}{\sqrt{\frac{si{n}^2\theta +co{s}^2\theta }{si{n}^2\theta .co{s}^2\theta }}}$

$=\frac{k.sin\theta cos\theta }{1}[\because si{n}^2\theta +co{s}^2\theta =1]$

$=\frac{k}{2}\times 2sin\theta cos\theta$

$=\frac{k}{2}sin2\theta [\because 2sin\theta cos\theta =sin2\theta ]$

$\therefore 2q=ksin2\theta$ . . .(ii)

On squaring Eqs. (i) and (ii) and then adding, we get

$p^2+4q^2=k^{2}co{s}^{2}2\theta +k^{2}si{n}^{2}2\theta$

$\Rightarrow p^2+q^2=k^2(co{s}^{2}2\theta +si{n}^{2}2\theta )$

$\therefore p^2+4q^2=k^2[\because co{s}^2\theta +si{n}^2\theta =1]$