Question

If $p$ and $q$ are the lengths of the perpendiculars from the origin on the lines, $x\text{cosec}\alpha -y\sec \alpha =k\cot 2\alpha$ and $x\sin \alpha +y\cos \alpha =k\sin 2\alpha$ respectively, then $k^2$ is equal to :

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

Answer 1

The correct option is C $4p^2+q^2$

Given lines are
$x\text{cosec}\alpha -y\sec \alpha =k\cot 2\alpha$ and $x\sin \alpha +y\cos \alpha =k\sin 2\alpha$
Distance from origin are
$p=\left|\frac{k\cot 2\alpha }{\sqrt{{\sec }^2\alpha +{\text{cosec}}^2\alpha }}\right|\Rightarrow p=\left|\frac{k\cot 2\alpha }{\sqrt{\frac{{\sin }^2\alpha +{\cos }^2\alpha }{{\sin }^2\alpha {\cos }^2\alpha }}}\right|\Rightarrow p=|k\cot 2\alpha ||\sin \alpha \cos \alpha |\Rightarrow p=|k\cot 2\alpha \times \frac{\sin 2\alpha }{2}|\Rightarrow p=\left|k\frac{\cos 2\alpha }{2}\right|\Rightarrow p^2=\frac{k^2}{4}{\cos }^{2}2\alpha \Rightarrow 4p^2=k^2{\cos }^{2}2\alpha \cdots \left(1\right)$

And
$q=\left|\frac{k\sin 2\alpha }{\sqrt{{\sin }^2\alpha +{\cos }^2\alpha }}\right|\Rightarrow q^2=k^2{\sin }^{2}2\alpha \cdots \left(2\right)$
Adding equation $\left(1\right)$ and $\left(2\right)$, we get
$4p^2+q^2=k^2$