मराठी

The normal of the curve x = a(cos θ + θ sin θ) and y = a(sin θ – θ cos θ), at any point θ, is such that ______.

Advertisements
Advertisements

प्रश्न

The normal of the curve x = a(cos θ + θ sin θ) and y = a(sin θ – θ cos θ), at any point θ, is such that ______.

पर्याय

  • it makes a constant angle with X-axis

  • it passes through the origin

  • it is parallel to Y-axis

  • it is at a constant distance from the origin

MCQ
रिकाम्या जागा भरा
Advertisements

उत्तर

The normal of the curve x = a(cos θ + θ sin θ) and y = a(sin θ – θ cos θ), at any point θ, is such that it is at a constant distance from the origin.

Explanation:

y = a(sin θ – θ cos θ), x = a(cos θ + θ sin θ) 

∴ `("d"y)/("d"theta)` = a(cos θ – cos θ + θ sin θ) = a θ sin θ and `("d"x)/("d"theta)` = a(– sin θ  + sin θ  + θ  cos θ) = a θ  cos θ 

∴ `("d"y)/("d"x) = (("d"y)/("d"theta))/(("d"x)/("d"theta)) = ("a"theta sin theta)/("a"theta costheta)` = tan θ

∴ Slope of the normal = `(-1)/(tan theta)` = – cot θ

∴ Equation of the normal is y – a sin θ + a θ cos θ

= `- costheta/sintheta (x - "a" cos  theta - "a"   theta sin theta)`

⇒ y sin θ – a sin2θ + aθ sin θ cos θ = – x cos θ + a cos2θ + aθ sin θ cos θ

⇒  x cos θ + y sin θ = a(sin2θ + cos2θ)

⇒ x cos θ + y sin θ = a

∴ Distance from origin = `|(-"a")/sqrt(sin^2theta + cos^2theta)|`

= a

= constant

shaalaa.com
Application of Derivative in Geometry
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×