The perpendicular distance from the origin to the normal drawn at any point on the curve x=a( cos θ+θ sin θ), y=a( sin θ-θ cos θ) is

Question

The perpendicular distance from the origin to the normal drawn at any point on the curve x=a( cos θ+θ sin θ), y=a( sin θ-θ cos θ) is (A) a θ (B) a2 (C) a (D) (a/θ)

Tardigrade Admin · Accepted Answer

Given, x=a( cos θ+θ sin θ) and y=a( sin θ-θ cos θ) Now, (d y/d x)=(d y/d θ) ⋅ (d θ/d x)= tan θ= Slope of tangent ∴ (d x/d θ)=(d(a cos θ+a θ sin θ)/d θ) =(d(a cos θ)/d θ)+(d(a θ sin θ)/d θ) =a(- sin θ)+a (d(θ sin θ)/d θ) =-a sin θ+a[ sin θ (d(θ)/d θ)+θ (d/d θ)( sin θ)] =-a sin θ+a sin θ+a θ cos θ=a θ cos θ Now, (d y/d θ)=(d(a sin θ-a θ cos θ)/d θ) =(d(a sin θ)/d θ)-(d(a θ cos θ)/d θ) =a cos θ-a (d(θ cos θ)/d θ) =a cos θ-a[ cos θ (d(θ)/d θ)+θ (d/d θ)( cos θ)] =a cos θ-a cos θ+a θ sin θ=a θ sin θ Now, (d y/d x)=(a θ sin θ/a θ cos θ)= tan θ ∴ Required equation of Normal (y-a sin θ+a θ cos θ) =(- cos θ/ sin θ)(x-a cos θ-a θ sin θ) ⇒ sin θ(y-a sin θ+a θ cos θ) =- cos θ(x-a cos θ-a θ sin θ) ⇒ y sin θ-a sin 2 θ+x cos θ-a cos 2 θ =a θ sin θ cos θ-a θ sin θ cos θ ⇒ y sin θ+x cos θ-a(1)=0 ⇒ x cos θ+y sin θ-a=0 So, perpendicular distance from origin, d=(| cos θ(0)+ sin θ ⋅(0)-a|/√ cos 2 θ+ sin 2 θ) =(|0+0-a|/√1)=(|-a|/1) ⇒ d=a

Q. The perpendicular distance from the origin to the normal drawn at any point on the curve
$x = a (cos θ + θ sin θ), y = a (sin θ - θ cos θ)$ is

$a θ$

$a^{2}$

$a$

$\frac{a}{θ}$

Q. The perpendicular distance from the origin to the normal drawn at any point on the curve x=a(cosθ+θsinθ),y=a(sinθ−θcosθ) is

aθ

a2

a

θa​

Q. The perpendicular distance from the origin to the normal drawn at any point on the curve
$x = a (cos θ + θ sin θ), y = a (sin θ - θ cos θ)$ is

$a θ$

$a^{2}$

$a$

$\frac{a}{θ}$