Question

If p and p' the lengths of perpendiculars from origin to the lines
x sec titha - y cosec titha= a and x cos titha - y sin titha= a cos 2titha, then 4p^2 +p'^2=
(a) 4a^2
(b) 2a^2
(c)a^2 (d) none of these​

Answer 1

Answer:

$\bold\red{(c){a}^{2}}$

Step-by-step explanation:

We know that,

Length of perpendicular from origin (0,0,0) on a line ax + by = c is Given by,

$perpeniclar \: length = | \frac{c}{ \sqrt{ {a}^{2} + {b}^{2} } } |$

Therefore,

we get,

$p = \frac{a}{ \sqrt{ {sec}^{2} \theta + {cosec}^{2} \theta} } = a \: sin\theta \: cos \theta \\ \\ = > {p}^{2} = {a}^{2} {sin}^{2} \theta \: {cos}^{2} \theta$

Similarly,

we get,

$p' = \frac{a \: cos2 \theta}{ \sqrt{ {sin}^{2} \theta + {cos}^{2} \theta } } = a \: cos 2\theta \\ \\ = > {p'}^{2} = {a}^{2} {cos}^{2} 2 \theta$

Therefore,

we get,

$4 {p}^{2} + {p'}^{2} = 4 {a}^{2} {sin}^{2} \theta \: {cos}^{2} \theta + {a}^{2} \: {cos}^{2}2 \theta \\ \\ = {(2a \: sin \theta \: cos \theta)}^{2} + {a}^{2} {cos}^{2} 2 \theta \\ \\ = {(a \: sin2 \theta)}^{2} + {a}^{2} {cos}^{2} 2 \theta \\ \\ = {a}^{2} {sin}^{2} 2 \theta + {a}^{2} {cos}^{2}2 \theta \\ \\ = {a}^{2} ( {sin}^{2}2 \theta + {cos}^{2} 2 \theta) \\ \\ = {a}^{2} \times 1 \\ \\ = {a}^{2}$

Hence, the correct option is

$\bold{(c){a}^{2}}$