Question

If $p$ and $p'$ be the distances of origin from the lines $\text{x sec }\\!\\!\alpha\\!\\!\text{ +y cosec }\\!\\!\alpha\\!\\!\text{ =k}$ and $\text{x cos}\alpha \text{-y sin}\alpha \text{=k cos2}\alpha$ then $4{{p}^{2}}+p{{'}^{2}}$

& \text{1) k} \\\ & \text{2) 2k} \\\ & \text{3) }{{\text{k}}^{\text{2}}} \\\ & \text{4) 2}{{\text{k}}^{\text{2}}} \\\ \end{aligned}$$

Answer 1

Hint : Firstly we will simplify the given equations and then by the distance between two parallel lines formula we may calculate the values in terms of $\text{p and p }\\!\\!'\\!\\!\text{ }$ and then we will put the values of $\text{p and p }\\!\\!'\\!\\!\text{ }$ in the given equation to check which option is correct in the above options.

Complete step-by-step solution:
As we already know parallel lines are those lines that never intersect each other even if you extend them. Parallel lines are always drawn on the same plane and they are always equidistant from each other.
Lines which meet or appear to meet when extended are called intersecting lines and the point where they meet is called the point of intersection.
When parallel lines get crossed by each other they are called transversal.
We know that slopes of two parallel lines are equal. Therefore, two parallel lines can be taken in the form
$y=mx+{{c}_{1}}$ say line$\text{1}$
and $y=mx+{{c}_{2}}$ say line $\text{2}$
line$\text{1}$ will intersect $x-axis$ at the point $(\dfrac{-{{c}_{1}}}{m},0)$
distance between the two lines is equal to the length of the perpendicular from a point to the line $\text{2}$
therefore the distance between the two lines will be given by $d=\dfrac{\left| {{c}_{1}}-{{c}_{2}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$
distance between the points $({{x}_{1}},{{y}_{1}})$ and a line is $d=\dfrac{\left| {{a}_{1}}{{x}_{1}}+{{b}_{1}}{{y}_{1}}+{{c}_{1}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$ where ${{a}_{1}}$ and ${{b}_{1}}$ are the coefficients of variable $x,y$ in the equation of line
the equation of line is ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$
Now according to the question:
$\text{x sec }\\!\\!\alpha\\!\\!\text{ +y cosec }\\!\\!\alpha\\!\\!\text{ =k}$
Making the equation in form of $a{{x}_{1}}+b{{y}_{1}}+{{c}_{1}}$ hence, $\text{x sec }\\!\\!\alpha\\!\\!\text{ +y cosec }\\!\\!\alpha\\!\\!\text{ -k=0}$
The distance of line $\text{x sec }\\!\\!\alpha\\!\\!\text{ +y cosec }\\!\\!\alpha\\!\\!\text{ -k=0}$of the above line from the origin $(0,0)$is given by
$p=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+{{c}_{1}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$
Where origin are $(0,0)$ means $x=0,y=0$
$p=\dfrac{\left| \text{0 }\\!\\!\times\\!\\!\text{ sec }\\!\\!\alpha\\!\\!\text{ +0 }\\!\\!\times\\!\\!\text{ cosec }\\!\\!\alpha\\!\\!\text{ -k} \right|}{\sqrt{\text{se}{{\text{c}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ +cose}{{\text{c}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ }}}$
$p=\dfrac{\left| -k \right|}{\sqrt{\text{se}{{\text{c}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ +cose}{{\text{c}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ }}}$
Making the equation in form of $a{{x}_{1}}+b{{y}_{1}}+{{c}_{1}}$ hence, $\text{x cos}\alpha \text{-y sin}\alpha -\text{k cos2}\alpha =0$
The distance of line $\text{x cos}\alpha \text{-y sin}\alpha -\text{k cos2}\alpha =0$ of the above line from the origin $(0,0)$is given by
Now calculating for $p'$
$p'=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+{{c}_{1}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$
Where origin are $(0,0)$ means $x=0,y=0$
$p'=\dfrac{\left| \text{0 }\\!\\!\times\\!\\!\text{ cos }\\!\\!\alpha\\!\\!\text{ -0 }\\!\\!\times\\!\\!\text{ sin }\\!\\!\alpha\\!\\!\text{ -kcos2}\alpha \right|}{\sqrt{{{\cos }^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ +si}{{\text{n}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ }}}$
$p'=\dfrac{\left| -k\cos 2\alpha \right|}{\sqrt{{{\cos }^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ +si}{{\text{n}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ }}}$
Now according to the question:
$4{{p}^{2}}+p{{'}^{2}}$
Putting the values of $p$ and $p'$ in $4{{p}^{2}}+p{{'}^{2}}$
$\dfrac{4{{k}^{2}}}{\text{se}{{\text{c}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ +cose}{{\text{c}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ }}+\dfrac{{{k}^{2}}\cdot {{\cos }^{2}}2\alpha }{{{\cos }^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ +si}{{\text{n}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ }}$
Taking ${{k}^{2}}$ in common and putting ${{\cos }^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ +si}{{\text{n}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ =1}$ we get
${{k}^{2}}[\dfrac{4}{\text{se}{{\text{c}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ +cose}{{\text{c}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ }}+{{\cos }^{2}}2\alpha ]$
Now by the formula $\cos 2\alpha ={{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha$
${{k}^{2}}[\dfrac{4}{\text{se}{{\text{c}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ +cose}{{\text{c}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ }}+{{({{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha )}^{2}}]$
${{k}^{2}}[\dfrac{4}{\dfrac{\text{1}}{{{\cos }^{2}}\alpha }\text{+}\dfrac{\text{1}}{{{\sin }^{2}}\alpha }}+{{({{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha )}^{2}}]$
${{k}^{2}}[\dfrac{4{{\sin }^{2}}\alpha \cdot {{\cos }^{2}}\alpha }{{{\cos }^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ +si}{{\text{n}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ }}+{{({{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha )}^{2}}]$
Putting ${{\cos }^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ +si}{{\text{n}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ =1}$ we get
${{k}^{2}}[4{{\sin }^{2}}\alpha \cdot {{\cos }^{2}}\alpha +{{\cos }^{4}}\alpha +{{\sin }^{4}}\alpha -2{{\sin }^{2}}\alpha \cdot {{\cos }^{2}}\alpha ]$
${{k}^{2}}[{{\cos }^{4}}\alpha +{{\sin }^{4}}\alpha +2{{\sin }^{2}}\alpha \cdot {{\cos }^{2}}\alpha ]$
According to the formula ${{a}^{2}}+{{b}^{2}}+2ab={{(a+b)}^{2}}$
${{k}^{2}}[{{({{\sin }^{2}}\alpha +co{{s}^{2}}\alpha )}^{2}}]$
putting ${{\cos }^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ +si}{{\text{n}}^{\text{2}}}\text{ }\\!\\!\alpha\\!\\!\text{ =1}$ we get
$={{k}^{2}}$

Hence by the above process we can prove that option$\text{(3)}$ is correct that $4{{p}^{2}}+p{{'}^{2}}={{k}^{2}}$

Note: Each line can have many parallel lines to it and parallel lines can be extended upto infinity. If you want to remember the sign of parallel lines you may take the example of an equals sign $\text{(=)}$or railway track, opposite walls and doors in a room , opposite sides of a blackboard.