Q83 / 160 · AP EAMCET 2013 · Mathematics
If $p$ and $q$ are the perpendicular distances from the origin to the straight lines $x \sec \theta-y \operatorname{cosec} \theta=a \quad$ and $x \cos \theta+y \sin \theta=a \cos 2 \theta$, then
- Correct answerA$4 p^2+q^2=a^2$
- B$p^2+q^2=a^2$
- C$p^2+2 q^2=a^2$
- D$4 p^2+q^2=2 a^2$
Solution
Given equations of straight lines are
$$
\begin{aligned}
x \sec \theta-y \operatorname{cosec} \theta & =a \\
x \cos \theta+y \sin \theta & =a \cos 2 \theta
\end{aligned}
$$
Also,
$p=$ Perpendicular distance from the origin to the
line (i)
$$
\begin{aligned}
& P=\frac{|0-0-a|}{\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}}=\frac{a \sin \theta \cdot \cos \theta}{\sqrt{1}} \\
&=a \sin \theta \cdot \cos \theta=\frac{a}{2} \sin 2 \theta \\
& \Rightarrow \quad 2 p=a \sin 2 \theta
\end{aligned}
$$
and $q=$ perpendicular distance from the origin to the line (ii)
$$
q=\frac{|0+0-a \cos 2 \theta|}{\sqrt{\cos ^2 \theta+\sin ^2 \theta}}=\frac{a \cos 2 \theta}{\sqrt{1}}=a \cos 2 \theta
$$
Now,
$$
\begin{aligned}
4 p^2+q^2 & =a^2 \sin ^2 2 \theta+a^2 \cos ^2 2 \theta \\
& =a^2\left(\sin ^2 2 \theta+\cos ^2 2 \theta\right) \\
& =a^2(1)=a^2
\end{aligned}
$$
Hence, the correct answer is option: 'A'.