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 $x \sec \theta+y \operatorname{cosec} \theta=k$ respectively, then

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

Answer 1

Answer: (B)

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