If p and q are the perpendicular distances from the origin to the straight lines x -y cosec- and x cos-y sin-cos 2 - thenA- p2-q2-a2B- 4 p 2- q 2- a 2C- p2-2 q2-a2D- 4 p2-q2-2 a2

The correct option is B 4p^2+q^2=a^2 p=1/2 |a sin 2θ|⇒ 2p=|a sin 2θ| q=|a cos 2θ| ⇒ 4p^2+q^2=a^2 ...

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Byju's Answer

Standard XII

Physics

Vector Addition

If p and q ar...

Question

If p and q are the perpendicular distances from the origin to the straight lines $x s e c θ - y c o s e c θ = α a n d x c o s θ + y s i n θ = α c o s 2 θ$ then

$4 p^{2} + q^{2} = a^{2}$

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$p^{2} + q^{2} = a^{2}$

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$p^{2} + 2 q^{2} = a^{2}$

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$4 p^{2} + q^{2} = 2 a^{2}$

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Solution

The correct option is A
$4 p^{2} + q^{2} = a^{2}$

$p = \frac{1}{2} | a s i n 2 θ | \Rightarrow 2 p = | a s i n 2 θ |$
$q = | a c o s 2 θ |$
$\Rightarrow 4 p^{2} + q^{2} = a^{2}$

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Similar questions

Q. If p₁ and p₂ are the lengths of the perpendiculars from the origin upon the lines x sec θ + y cosec θ = a and x cos θ − y sin θ = a cos 2 θ respectively, then
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Q. If

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4 p^{2} + q^{2}

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If p and q are the perpendicular distances from the origin to the straight lines x secθ−y cosecθ=α and x cosθ+y sinθ=αcos 2θ then

The correct option is A 4p2+q2=a2 p=12|a sin 2θ|⇒2p=|a sin 2θ| q=|a cos 2θ| ⇒4p2+q2=a2

If p and q are the perpendicular distances from the origin to the straight lines $x s e c θ - y c o s e c θ = α a n d x c o s θ + y s i n θ = α c o s 2 θ$ then

The correct option is A
$4 p^{2} + q^{2} = a^{2}$

$p = \frac{1}{2} | a s i n 2 θ | \Rightarrow 2 p = | a s i n 2 θ |$
$q = | a c o s 2 θ |$
$\Rightarrow 4 p^{2} + q^{2} = a^{2}$