Question

If p and q are the lengths of perpendiculars from the origin to the lines x cosᎾ +y sin Ꮎ= k cos 2Ꮎand x sec Ꮎ - ycosecᎾ= k, respectively, prove that 4p²+q²=k²​

Answer 1

Step-by-step explanation:

Lines are

$\blue{L_{1} \colon x \cos( \theta) + y \sin( \theta) - k \cos(2 \theta) = 0 } \\ \blue{ L_{2} \colon \: x \sec( \theta) - y \cosec( \theta) - k = 0 }\: \: \: \: \: \: \: \: \: \: \:$

$\large{\tt\:\red{p}\:\:and\:\: \red{q}\:\:are\:\:the\:\:lengths\:\:of\:\:perpendiculars}\\\large{\tt\:from\:\:origin\:\:to\:\:the\:\:lines\:\:\pink{L_{1}}\:\:and\:\:\pink{L_{2}}}$

So,

$\red{p} = \frac{ |(0) \cos( \theta) + (0) \sin( \theta) - k \cos( 2\theta) | }{ \sqrt{ \cos^{2} ( \theta) + \sin^{2} ( \theta) } }$

$\implies\red{p} = \frac{ | - k \cos( 2\theta) | }{ \sqrt{1} }$

$\implies\red{p} = k \cos( 2\theta)$

And,

$\red{q} = \frac{ |(0) \sec( \theta) - (0) \cosec( \theta) - k | }{ \sqrt{ \sec^{2} ( \theta) + \cosec^{2} ( \theta) } }$

$\implies \red{q} = \frac{ | - k | }{ \sqrt{ \dfrac{1}{\cos^{2} ( \theta)} + \dfrac{1}{\sin^{2} ( \theta) }} }$

$\implies \red{q} = \frac{ k }{ \sqrt{ \dfrac{ \sin^{2} ( \theta) +\cos^{2} ( \theta) }{\cos^{2} ( \theta)\sin^{2} ( \theta) } } }$

$\implies \red{q} = \frac{ k }{ \dfrac{\sqrt{ \sin^{2} ( \theta) +\cos^{2} ( \theta) }}{ \sin (\theta) \cos (\theta)} }$

$\implies \red{q} = \frac{ k \sin (\theta) \cos (\theta) }{\sqrt{ \sin^{2} ( \theta) +\cos^{2} ( \theta) }}$

$\implies \red{q} = \frac{ k \sin (\theta) \cos (\theta) }{\sqrt{ 1 }}$

$\implies \red{q} = k \sin (\theta) \cos (\theta)$

Now,

$4 {q}^{2} + {p}^{2} = 4 {k} ^{2} \sin^{2} ( \theta) \cos^{2} ( \theta) + {k}^{2} \cos^{2} (2 \theta)$

$\implies4 {q}^{2} + {p}^{2} = {k} ^{2}.4 \sin^{2} ( \theta) \cos^{2} ( \theta) + {k}^{2} \cos^{2} (2 \theta)$

$\implies4 {q}^{2} + {p}^{2} = {k} ^{2}. \{2 \sin ( \theta) \cos ( \theta) \}^{2} + {k}^{2} \cos^{2} (2 \theta)$

$\implies4 {q}^{2} + {p}^{2} = {k} ^{2}. \{\sin ( 2\theta) \}^{2} + {k}^{2} \cos^{2} (2 \theta)$

$\implies4 {q}^{2} + {p}^{2} = {k} ^{2}. \sin^{2} ( 2\theta) + {k}^{2} \cos^{2} (2 \theta)$

$\implies4 {q}^{2} + {p}^{2} = {k} ^{2} \{\sin^{2} ( 2\theta) + \cos^{2} (2 \theta) \}$

$\implies4 {q}^{2} + {p}^{2} = {k} ^{2} \{1 \}$

$\purple{ \sf\implies 4 {q}^{2} + {p}^{2} = {k} ^{2}}$