Question

If $p$ and $p\text{'}$ be the distances of origin from the lines$xsec\alpha +y\cos ec\alpha =k$ and$x\cos \alpha -y\sin \alpha =k\cos 2\alpha$, then $4p^2+p{\text{'}}^2$

• A. $k$
• B. $2k$
• C. $k^2$
• D. $2k^2$

Answer 1

The correct option is C

$k^2$

Step1: Given data and Perpendicular distance formula between a line and a point:

Given two lines are

$xsec\alpha +y\cos ec\alpha =k$ -----------(i)

$x\cos \alpha -y\sin \alpha =k\cos 2\alpha$ ------------(ii)

The distance between of a line $ax+by+c=0$ and a point $(x_1,y_1)$ is given by

$d=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b2}}\right|$ ------------(iii)

Step2: Calculate the value of $p$

Consider equation (i),

$\begin{array}{rcl}\therefore xsec\alpha +y\cos ec\alpha & =& k\\ & \Rightarrow & xsec\alpha +y\cos ec\alpha -k=0\\ & \Rightarrow & x\left(\frac{1}{\cos \alpha }\right)+y\left(\frac{1}{\sin \alpha }\right)-k=0\\ & \Rightarrow & x\sin \alpha +y\cos \alpha -k\sin \alpha \cos \alpha =0\end{array}$

The above equation is in the form of $ax+by+c=0$

where $a=\sin \alpha$, $b=\cos \alpha$, $c=-k\sin \left(\alpha \right)\cos \left(\alpha \right)$

Also, $p$ is the perpendicular distance from the origin$(0,0)$ to line ( i ),

Here $x_1=0$, $y_1=0$, $d=p$

By substituting the above values in (iii), we get

$\begin{array}{rcl}\therefore p& =& \left|\frac{\sin \alpha \left(0\right)+\cos \alpha \left(0\right)-k\sin \alpha \cos \alpha }{\sqrt[]{{\sin }^2\alpha +{\cos }^2\alpha }}\right|\\ & =& \left|\frac{-k\sin \left(\alpha \right)\cos \left(\alpha \right)}{1}\right|\\ & =& k\sin \left(\alpha \right)\cos \left(\alpha \right)\end{array}$ $\left[\because {\sin }^2\alpha +{\cos }^2\alpha =1\right]$

Step 2: Calculate the value of $p\text{'}$

Consider line (ii),

$x\cos \alpha -y\sin \alpha =k\cos 2\alpha$

$x\cos \alpha -y\sin \alpha -k\cos 2\alpha =0$

The above equation is in the form of $ax+by+c=0$

Where $a=\cos \alpha$, $b=-\sin \alpha$, $c=-k\cos 2\alpha$

Also, $p\text{'}$ is the perpendicular distance from the origin$(0,0)$ to the line (ii)

Here $x_1=0$, $y_1=0$, $d=p\text{'}$

By substituting the above values in (iii), we get

$\begin{array}{rcl}\therefore p\text{'}& =& \left|\frac{\cos \alpha \left(0\right)+(-\sin \alpha )\left(0\right)-k\cos 2\alpha }{\sqrt{{\sin }^2\alpha +{\cos }^2\alpha }}\right|\\ & =& \left|\frac{-k\cos 2\alpha }{1}\right|\\ & =& k\cos 2\alpha \end{array}$ $\left[\because {\sin }^2\alpha +{\cos }^2\alpha =1\right]$

Step-3: Substitute the values of $p$ and $p\text{'}$in $4p^2+p{\text{'}}^2$

We have to find the value of $4p^2+p{\text{'}}^2$

Substitute the values of $p$ and $p\text{'}$in $4p^2+p{\text{'}}^2$ then we get

$\begin{array}{rcl}\therefore 4{\left(k\sin \alpha \cos \alpha \right)}^2+{\left(k\cos 2\alpha \right)}^2& =& 4k^2{\sin }^2\alpha {\cos }^2\alpha +k^2{\cos }^{2}2\alpha \\ & =& k^2\left(4{\sin }^2\alpha {\cos }^2\alpha +{\cos }^{2}2\alpha \right)\end{array}$

Substitute ${\cos }^{2}2\alpha ={\cos }^2\alpha -{\sin }^2\alpha$

$\begin{array}{rcl}\therefore 4{\left(k\sin \alpha \cos \alpha \right)}^2+{\left(k\cos 2\alpha \right)}^2& =& k^2\left(4{\sin }^2\alpha {\cos }^2\alpha +{\left({\cos }^2\alpha -{\sin }^2\alpha \right)}^2\right)\\ & =& k^2\left(4{\sin }^2\alpha {\cos }^2\alpha +({\cos }^2\alpha +{\sin }^2\alpha -2{\cos }^2\alpha {\sin }^2\alpha \right))\\ & =& k^2\left({\cos }^2\alpha +{\sin }^2\alpha +2{\cos }^2\alpha {\sin }^2\alpha \right)\\ & =& k^2{\left({\cos }^2\alpha +{\sin }^2\alpha \right)}^2\\ & =& k^2\end{array}$

Clearly $4p^2+p{\text{'}}^2=k^2$

Hence, option (C) is correct.