If P_ 1 and P_ 2 be the length of perpendiculars from the origin upon the straight lines x θ+y cosec θ=a and x θ-y θ=a 2 θ respectively, then the value of 4 P_ 1 ^ 2 +P_ 2 ^ 2

Correct answer: A. a^ 2 We have P_ 1 = length of perpendicular from (0,0) on x θ+y cosec θ=a aligned i.e. P_ 1 &=| a ^ 2 θ+ e c^ 2 θ |=|a θ θ| &=|(a)/(2) 2 θ| or 2 P_ 1 =|a 2 θ| aligned P_ 2 = Length of the perpendicular from (0,0) on x θ-y θ=a 2 _ 2 =| a 2 θ ^ 2 θ+ ^ 2 θ |=|a 2 θ| Now, 4 P_ 1 ^ 2 +P_ 2 ^ 2 =a^ 2 ^ 2 2 θ+a^ 2 ^ 2 2 θ=a^ 2

BITSAT 2014

Exam: BITSAT
Year: 2014
Subject: Mathematics
Chapter: Straight Lines

BITSAT 2014 · Mathematics · Straight Lines

If P_ 1 and P_ 2 be the length of perpendiculars from the origin...

P_{1}

and

P_{2}

be the length of perpendiculars from the origin upon the straight lines

x \sec \theta+y \operatorname{cosec} \theta=a

and

x \cos \theta-y \sin \theta=a \cos 2 \theta

respectively, then the value of

4 \mathrm{P}_{1}^{2}+\mathrm{P}_{2}^{2}

Options

A $a^{2}$
B $2 a^{2}$
C $a^{2} / 2$
D $3 a^{2}$

Correct answer

A. $a^{2}$

Step-by-step solution

We have

P_{1}=

length of perpendicular from (0,0) on

x \sec \theta+y \operatorname{cosec} \theta=a

\begin{aligned} \text{ i.e. } P_{1} &=\left|\frac{a}{\sqrt{\sec ^{2} \theta+\cos e c^{2} \theta}}\right|=|a \sin \theta \cos \theta| \\ &=\left|\frac{a}{2} \sin 2 \theta\right| \text{ or } 2 P_{1}=|a \sin 2 \theta| \end{aligned}

\mathrm{P}_{2}=

Length of the perpendicular from (0,0) on

x \cos \theta-y \sin \theta=a \cos 2 \thetaP_{2}=\left|\frac{a \cos 2 \theta}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}\right|=|a \cos 2 \theta|

Now,

4 P_{1}^{2}+P_{2}^{2}=a^{2} \sin ^{2} 2 \theta+a^{2} \cos ^{2} 2 \theta=a^{2}

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