Question

If $p$ and $q$ are lengths of perpendiculars from origin to the lines $x\cos \theta -y\sin \theta =k\cos 2\theta$ and $x\sec \theta +y\mathrm{cosec}\theta =k$ respectively, then

• A. $4q^2+p^2=k^2$
• B. $p^2+4q^2=k^2$
• C. $p^2+q^2=k^2$
• D. $4p^2+4q^2=k^2$

Answer 1

Answer: (B)

For figure (a), $x\cos \theta -y\sin \theta =k\cos 2\theta ....$(i)
For figure (b) $x\sec \theta +y\mathrm{cosec}\theta =k....$(ii)
$\therefore$ Perpendicular distance from $(0,0)$ to the Eq. (i) is
$p=\left|\frac{0-0-k\cos 2\theta }{\sqrt{{\cos }^2\theta +{\sin }^2\theta }}\right|=k\cos 2\theta ....$(iii)
$\left(\because d=\mid \frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right)$
Again, Eq. (ii) can be written as
$\frac{x}{\cos \theta }+\frac{y}{\sin \theta }=k$
$\Rightarrow x\sin \theta +y\cos \theta =k\cos \theta \sin \theta$
$\Rightarrow x\sin \theta +y\cos \theta =\frac{k}{2}(2\cos \theta \sin \theta )$
$\Rightarrow x\sin \theta +y\cos \theta =\frac{k}{2}\sin 2\theta$
$\Rightarrow x\sin \theta +y\cos \theta -\frac{k}{2}\sin 2\theta =\mathrm{0....}$(iv)
Now, perpendicular distance from $(0,0)$ to Eq. (iv) is
$q=\left|\frac{0+0-\frac{k}{2}\sin 2\theta }{\sqrt{{\sin }^2\theta +{\cos }^2\theta }}\right|$
$\left(\because a=\sin \theta ,b=\cos \theta ,c=-\frac{k}{2}\sin 2\theta ,x_1=0,y_1=0\right)$
$\Rightarrow q=\frac{k}{2}\sin 2\theta \left(\because {\sin }^2\theta +{\cos }^2\theta =1\right)$
$\Rightarrow 2q=k\sin 2\theta ....$(v)
On squaring and adding Eqs. (iii) and (v), we get
$p^2+4q^2=k^2{\cos }^{2}2\theta +k^2{\sin }^{2}2\theta$
$=k^2\left({\cos }^{2}2\theta +{\sin }^{2}2\theta \right)$
$\therefore p^2+4q^2=k^2\left(\because {\sin }^2\theta +{\cos }^2\theta =1\right)$