Question

The normal of the curve given by the equation x = a(sinθ + cosθ), y = a(sinθ – cosθ) at the point θ is ______.

Options

• (x + y)cosθ + (x – y)sinθ = 0

• (x + y)cosθ + (x – y)sinθ = a

• (x + y)cosθ – (x – y)sinθ = 0

• (x + y)cosθ – (x – y)sinθ = a

Answer 1

The normal of the curve given by the equation x = a(sinθ + cosθ), y = a(sinθ – cosθ) at the point θ is (x + y)cosθ – (x – y)sinθ = 0.

Explanation:

x = a(sinθ + cosθ), y = a(sinθ – cosθ)

Differentiating above functions w.r.t.θ, we get

`(dx)/(dθ)` = a(cosθ – sinθ)

`(dy)/(dθ)` = a(cosθ + sinθ)

`(dy/(dθ))/(dx/(dθ)) = (dy)/(dx)`

= `(a(cosθ + sinθ))/(a(cosθ - sinθ))`

`(dy)/(dx) = (cosθ + sinθ)/(cosθ - sinθ)`

Slope of normal = `-1/((dy/dx))`

= `-((cosθ - sinθ))/((cosθ + sinθ))`

= `((sinθ - cosθ))/((sinθ + cosθ))`

Equation of normal

y – a(sinθ – cosθ) = `(sinθ - cosθ)/(sinθ + cosθ)` [x – a(sinθ + cosθ)]

y(sinθ + cosθ) – a(sinθ – cos²θ) = x(sinθ – cosθ) – a(sin²θ – cos²θ)

⇒ y(sinθ + cosθ) = x(sinθ – cosθ)

⇒ (y – x)sinθ + (y + x)cosθ = 0

⇒ (x + y)cosθ – (x – y)sinθ = 0