Question

If $P$ and $P^{\prime }$ are the perpendiculars from the origin, upon the straight lines $x\sec \theta +y\csc \theta =a$ and $x\cos \theta -y\sin \theta =a\cos 2\theta ,$ then the value of $4P^2+{P^{\prime }}^2$ is

• A. $-12a^2$
• B. $-5a^2$
• C. $a^2$
• D. $5a^2$

Answer 1

The correct option is C $a^2$

Given : Equations of lines

$x\sec \theta +y\csc \theta =a$$\Rightarrow x\sec \theta +y\csc \theta -a=0$ ...(1)

and $x\cos \theta -y\sin \theta =a\cos 2\theta$$\Rightarrow x\cos \theta -ysin\theta -a\cos 2\theta =0$ ...(2)

We know that the coordinates of the origin $(x_1,y_1)=(0,0)$.

We also know that the standard equation of the line is $ax+by+c=0$.

Comparing Eqs. (1) and (2) with the standard equation, we get

$a_1=\sec \theta ,b_1=\csc \theta ,c_1=-a,a_2=\cos \theta ,$$b_2=-\sin \theta$ and $c_2=-a\cos 2\theta$

Therefore,the length of the perpendicular from the origin to the line (1) is given by

$\left(p\right)=\frac{|a{x}_1+b{y}_1+c_1|}{\sqrt{a^2+b^2}}=\frac{|\sec \theta \times \left(0\right)+\csc \theta \times \left(0\right)-a|}{\sqrt{{\sec }^2\theta +{\csc }^2\theta }}$

$=\frac{a\cos \theta \sin \theta }{\sqrt{{\sin }^2\theta +{\cos }^2\theta }}=a\sin \theta \cos \theta$

Similarly, the length of the perpendicular from the origin to the line (2) is given by

$\left(p^{\prime }\right)=\frac{|a_2{x}_1+b_2{y}_1+c_2|}{\sqrt{a^2+b^2}}$

$=\frac{\left|\cos \theta \times \left(0\right)-\sin \theta \times \left(0\right)-a\cos 2\theta \right|}{\sqrt{{\cos }^2\theta +{\sin }^2\theta }}=a\cos 2\theta$

Therefore,$4p^2+p^{\prime 2}=4a^2{\sin }^2\theta {\cos }^2\theta +a^2{\cos }^{2}2\theta$

$=a^2{\left(2\sin \theta \cos \theta \right)}^2+a^2{\cos }^{2}2\theta =a^2({\sin }^{2}2\theta +{\cos }^{2}2\theta )=a^2$