If p and q are the lengths of perpendiculars from the origin to the lines
Question:

If $p$ and $q$ are the lengths of perpendiculars from the origin to the lines $x \cos \theta-y \sin \theta=k \cos 2 \theta$ and $x \sec \theta+y \operatorname{cosec} \theta=k$, respectively, prove that $p^{2}+4 q^{2}=k^{2}$

Solution:

The equations of given lines are

$x \cos \theta-y \sin \theta=k \cos 2 \theta \ldots$ (1)

$x \sec \theta+y \operatorname{cosec} \theta=k$ (2)

The perpendicular distance $(d)$ of a line $A x+B y+C=0$ from a point $\left(x_{1}, y_{1}\right)$ is given by $d=\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}$.

On comparing equation (1) to the general equation of line i.e., $A x+B y+C=0$, we obtain $A=\cos \theta, B=-\sin \theta$, and $C=-k \cos 2 \theta$.

It is given that is the length of the perpendicular from (0, 0) to line (1).

$\therefore p=\frac{|A(0)+B(0)+C|}{\sqrt{A^{2}+B^{2}}}=\frac{|C|}{\sqrt{A^{2}+B^{2}}}=\frac{|-k \cos 2 \theta|}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}=|-k \cos 2 \theta|$ (3)

On comparing equation (2) to the general equation of line i.e., $A x+B y+C=0$, we obtain $A=\sec \theta, B=\operatorname{cosec} \theta$, and $C=-k$.

It is given that is the length of the perpendicular from (0, 0) to line (2).

$\therefore q=\frac{|A(0)+B(0)+C|}{\sqrt{A^{2}+B^{2}}}=\frac{|C|}{\sqrt{A^{2}+B^{2}}}=\frac{|-k|}{\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}}$ (4)

From $(3)$ and $(4)$, we have

$p^{2}+4 q^{2}=(|-k \cos 2 \theta|)^{2}+4\left(\frac{|-k|}{\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}}\right)^{2}$

$=k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{\left(\sec ^{2} \theta+\operatorname{cosec}^{2} \theta\right)}$

$=k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{\left(\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}\right)}$

$=k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{\left(\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta \cos ^{2} \theta}\right)}$

$=k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{\left(\frac{1}{\sin ^{2} \theta \cos ^{2} \theta}\right)}$

$=k^{2} \cos ^{2} 2 \theta+4 k^{2} \sin ^{2} \theta \cos ^{2} \theta$

$=k^{2} \cos ^{2} 2 \theta+k^{2}(2 \sin \theta \cos \theta)^{2}$

$=k^{2} \cos ^{2} 2 \theta+k^{2} \sin ^{2} 2 \theta$

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

$=k^{2}$

Hence, we proved that $p^{2}+4 q^{2}=k^{2}$.

 

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