SATHEE: Application of Derivatives 1 Question 13

Application of Derivatives 1 Question 13

####13. The normal to the curve $x = a (\cos θ + θ \sin θ)$ , $y = a (\sin θ - θ \cos θ)$ at any point ’ $θ$ ’ is such that

(1983, 1M)

(a) it makes a constant angle with the $X$ -axis

(b) it passes through the origin

(d) None of the above

Objective Questions II

(One or more than one correct option)

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Answer:

Correct Answer: 13. $x + 2 y = \frac{π}{2}$ and $x + 2 y = \frac{- 3 π}{2}$

Solution:

Given, $x = a (\cos θ + θ \sin θ)$

and $y = a (\sin θ - θ \cos θ)$

$∴ \frac{d x}{d θ} = a (- \sin θ + \sin θ + θ \cos θ) = a θ \cos θ$

and $\frac{d y}{d θ} = a (\cos θ - \cos θ + θ \sin θ)$

$\frac{d y}{d θ} = a θ \sin θ \Rightarrow \frac{d y}{d x} = \tan θ$

Thus, equation of normal is

$\frac{y - a (\sin θ - θ \cos θ)}{x - a (\cos θ + θ \sin θ)} = \frac{- \cos θ}{\sin θ}$

$\Rightarrow - x \cos θ + a θ \sin θ \cos θ + a \cos^{2} θ$ $= y \sin θ + θ a \sin θ \cos θ - a \sin^{2} θ$

$\Rightarrow x \cos θ + y \sin θ = a$

whose distance from origin is

$\frac{| 0 + 0 - a |}{\sqrt{\cos^{2} θ + \sin^{2} θ}} = a$

Application of Derivatives 1 Question 13

Objective Questions II

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

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Sample Mock Test

Mentorship

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Application of Derivatives 1 Question 13

Objective Questions II

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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