Question

If p and q are perpendicular distances from origin to the straight lines $x\sec \theta -y\csc \theta =a$ and $x\cos \theta +y\sin \theta =a\cos 2\theta$, then which of the following is correct ?
(a) $4{{p}^{2}}+{{q}^{2}}={{a}^{2}}$
(b) ${{p}^{2}}+{{q}^{2}}={{a}^{2}}$
(c) ${{p}^{2}}+2{{q}^{2}}={{a}^{2}}$
(d) $4{{p}^{2}}+2{{q}^{2}}=2{{a}^{2}}$

Answer 1

Hint: Use the formula for calculating the perpendicular distance from a point $\left( {{x}_{1}},{{y}_{1}} \right)$ to the line $Ax+By+C=0$ i.e. given as $\left| \dfrac{A{{x}_{1}}+B{{y}_{1}}+C}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right|$. Convert $\sec \theta$ and $\csc \theta$ to $\dfrac{1}{\cos \theta }$ and $\dfrac{1}{\sin \theta }$ respectively and now try to eliminate $\theta$ and use ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.

Complete step-by-step solution -
We know the perpendicular distance of a line $Ax+By+C=0$ by a general point $\left( {{x}_{1}},{{y}_{1}} \right)$ can be given by the relation
Perpendicular distance$=\left| \dfrac{A{{x}_{1}}+B{{y}_{1}}+C}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right|$ ……………………………………(i)
So, we have two different lines $x\sec \theta -y\csc \theta =a$ and $x\cos \theta +y\sin \theta =a\cos 2\theta$ and perpendicular distance of each lines from the origin are given as p and q respectively.
Hence, we can calculate the perpendicular distance of each line by equation (i) and can equate to the given distances p and q.
So, perpendicular distance from line 1 from origin $\left( 0,0 \right)$ can be given as
$p=\left| \dfrac{0\left( \sec \theta \right)-0\left( \csc \theta \right)-a}{\sqrt{{{\left( \sec \theta \right)}^{2}}+{{\left( -\csc \theta \right)}^{2}}}} \right|$
$p=\left| \dfrac{-a}{\sqrt{{{\sec }^{2}}\theta +{{\csc }^{2}}\theta }} \right|$
On squaring both sides, we can remove the modulus sign as the square of any number will never be negative. Hence, we get
${{p}^{2}}={{\left( \dfrac{-a}{\sqrt{{{\sec }^{2}}\theta +{{\csc }^{2}}\theta }} \right)}^{2}}$
${{p}^{2}}=\dfrac{{{a}^{2}}}{{{\sec }^{2}}\theta +{{\csc }^{2}}\theta }$
Now, we can change $\sec \theta$ to $\cos \theta$ by relation of $\sec \theta =\dfrac{1}{\cos \theta }$ and $\csc \theta$ to $\sin \theta$ by relation $\csc \theta =\dfrac{1}{\sin \theta }$.
Hence, we can simplify ${{p}^{2}}$ as
${{p}^{2}}=\dfrac{{{a}^{2}}}{\left( \dfrac{1}{{{\cos }^{2}}\theta }+\dfrac{1}{{{\sin }^{2}}\theta } \right)}$
${{p}^{2}}=\dfrac{{{a}^{2}}}{\dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }}$
Now, we can replace ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta$ by 1 using the relation
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Hence, we get
${{p}^{2}}=\dfrac{{{a}^{2}}}{\dfrac{1}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }}$
${{p}^{2}}={{a}^{2}}{{\sin }^{2}}\theta {{\cos }^{2}}\theta$ ….....................................................(ii)
Now, similarly we can get perpendicular distance of line $x\cos \theta +y\sin \theta =a\cos 2\theta$ from origin $\left( 0,0 \right)$ by equation (i) and equating it to ‘q’. Hence, we get
$q=\left| \dfrac{0\left( \cos \theta \right)+0\left( \sin \theta \right)-a\cos 2\theta }{\sqrt{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }} \right|$
$\Rightarrow q=\left| \dfrac{-a\cos 2\theta }{\sqrt{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }} \right|$
Now, on squaring both sides, we get
${{q}^{2}}=\dfrac{{{\left( -a\cos 2\theta \right)}^{2}}}{\left( \sqrt{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta } \right)}$
${{q}^{2}}=\dfrac{{{a}^{2}}{{\cos }^{2}}2\theta }{\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)}$
Now, we can replace ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta$ by 1 using the trigonometric identity given as ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$
Hence, we can get value of ${{q}^{2}}$ as
${{q}^{2}}={{a}^{2}}{{\cos }^{2}}2\theta$ …............................(iii)
Now, we know the trigonometric identity of $\cos 2\theta$, given as
$\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta$ ………………………………….(iv)
Hence, we can replace $\cos 2\theta$ by ${{\cos }^{2}}\theta -{{\sin }^{2}}\theta$ using equation (iv) in equation (iii) ; So, we get
${{q}^{2}}={{a}^{2}}{{\left( \cos 2\theta \right)}^{2}}$
${{q}^{2}}={{a}^{2}}\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)$ …………………………..(v)
Hence, we can rewrite equation (v) by using algebraic identity ${{\left( a-b \right)}^{2}}$ given as
${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
So, we get the value of ${{q}^{2}}$ from equation (v) using the above algebraic identity. Hence, we get
${{q}^{2}}={{a}^{2}}\left( {{\cos }^{4}}\theta +{{\sin }^{4}}\theta -2{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)$
Now, add and subtract $2{{\sin }^{2}}\theta {{\cos }^{2}}\theta$ in the bracket of above equation. So, we get
${{q}^{2}}={{a}^{2}}\left( {{\cos }^{4}}\theta +{{\sin }^{4}}\theta -2{{\sin }^{2}}\theta {{\cos }^{2}}\theta +2{{\sin }^{2}}\theta {{\cos }^{2}}\theta -2{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)$
$\Rightarrow {{q}^{2}}={{a}^{2}}\left( {{\cos }^{4}}\theta +{{\sin }^{4}}\theta +2{{\sin }^{2}}\theta {{\cos }^{2}}\theta -4{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)$
$\Rightarrow {{q}^{2}}={{a}^{2}}\left[ {{\left( {{\sin }^{2}}\theta \right)}^{2}}+{{\left( {{\cos }^{2}}\theta \right)}^{2}}+2\times {{\sin }^{2}}\theta \times {{\cos }^{2}}\theta -4{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right]$
Now, we can use algebraic identity of ${{\left( a+b \right)}^{2}}$ to simplify the above equation, which is given as
${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
Hence, we get value of ${{q}^{2}}$ as
${{q}^{2}}={{a}^{2}}\left( {{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}^{2}}-4{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)$
Now, put ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta$ as 1 using relation
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Hence, we get
${{q}^{2}}={{a}^{2}}\left( 1-4{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)$ ……………………………………….(vi)
Now, we can get value of ${{\sin }^{2}}\theta {{\cos }^{2}}\theta$ from the equation (ii),
${{p}^{2}}={{a}^{2}}{{\sin }^{2}}\theta {{\cos }^{2}}\theta$
$\Rightarrow {{\sin }^{2}}\theta {{\cos }^{2}}\theta =\dfrac{{{p}^{2}}}{{{a}^{2}}}$ ………………………………………..(vii)
Hence, we can replace ${{\sin }^{2}}\theta {{\cos }^{2}}\theta$ by $\dfrac{{{p}^{2}}}{{{a}^{2}}}$ in the equation (vi). Hence, we get
${{q}^{2}}={{a}^{2}}\left( 1-\dfrac{4{{p}^{2}}}{{{a}^{2}}} \right)$
${{q}^{2}}={{a}^{2}}\left( \dfrac{{{a}^{2}}-4{{p}^{2}}}{{{a}^{2}}} \right)$
$\Rightarrow {{q}^{2}}={{a}^{2}}-4{{p}^{2}}$
$\Rightarrow {{q}^{2}}+4{{p}^{2}}={{a}^{2}}$
Hence, option (A) is correct.

Note: Don’t get confused with the modulus sign with the perpendicular distance formula. Squaring both the sides will remove the modulus function.
One can use ${{\left( a+b \right)}^{2}}={{\left( a-b \right)}^{2}}+4ab$ to simplify ${{\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)}^{2}}$. So, we get ${{\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)}^{2}}={{\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)}^{2}}+4{{\cos }^{2}}\theta {{\sin }^{2}}\theta$ .
Calculation is an important part of the question as well. So, be careful with it as well.